First off, I know that this problem is extremely long and can be intimidating. I have put a lot of work and thought into it.
In reference to what loup blanc has responded with, I am not dismissing his answer, I just do not understand it. Please either respond according to my what, or elaborate on a different approach with as much elaboration as I have put in please.
$DT_3(R)$ is the upper triangular matrix with the diagonal being the same element over the real field R, for example $\begin{bmatrix}a&b&c\\0&a&d\\0&0&a\end{bmatrix}$ where $a,b,c,d ∈R$. Based on this, we will show how the following conditions hold true. Determine all mappings s.t $DT_3(R) \to DT_3(R)$ and the following conditions are true:
$(1) d(A+B)=d(A)+d(B)$
$(2) d(αA)=αd(A)$ where $α$ is a scalar
$(3) d(AB)=Ad(B)+d(A)B$
Let's suppose $\begin{bmatrix}a&b&c\\0&a&d\\0&0&a\end{bmatrix}$ $\to \begin{bmatrix}f(a,b,c,d)&g(a,b,c,d)&h(a,b,c,d)\\0&f(a,b,c,d)&i(a,b,c,d)\\0&0&f(a,b,c,d)\end{bmatrix}$ where $f,g,h,i$ from $R^4 \to R$
Also, suppose $A=\begin{bmatrix}a&b&c\\0&a&d\\0&0&a\end{bmatrix}$ and $B=\begin{bmatrix}x&y&z\\0&x&v\\0&0&x\end{bmatrix}$
We will now work on finding the restrictions on f,g,h,i s.t. the above conditions, starting with condition $(1) d(A+B)=d(A)+d(B)$
therefore $d(A+B)=\begin{bmatrix}f(a+x,b+y,c+z,d+v)&g(a+x,b+y,c+z,d+v)&h(a+x,b+y,c+z,d+v)\\0&f(a+x,b+y,c+z,d+v)&i(a+x,b+y,c+z,d+v)\\0&0&f(a+x,b+y,c+z,d+v)\end{bmatrix}$
This is the left side of the equation. Now to show the right side
$d(A)+d(B)=\begin{bmatrix}f(a,b,c,d)+f(x,y,z,v)&g(a,b,c,d)+g(x,y,z,v)&h(a,b,c,d)+h(x,y,z,v)\\0&f(a,b,c,d)+f(x,y,z,v)&i(a,b,c,d)+i(x,y,z,v)\\0&0&f(a,b,c,d)+f(x,y,z,v)\end{bmatrix}$
Taking each of the elements in the above matrix and adding the respective elements together, we get the following equations:
$f(a+x,b+y,c+z,d+v)=f(a,b,c,d)+f( x,y,z,v)$ $g(a+x,b+y,c+z,d+v)=g(a,b,c,d)+g( x,y,z,v)$ $h(a+x,b+y,c+z,d+v)=h(a,b,c,d)+h( x,y,z,v)$ $i(a+x,b+y,c+z,d+v)=i(a,b,c,d)+i( x,y,z,v)$
Therefor f,g,h,i preservee addition.
Condition (2) states $d(αA)=αd(A)$ where $α$ is a scalar. We will take matrix A from above and multiply it by a scalar. We will show the left side of the equal sign first, then we will show the right side.
$d(αA)=\begin{bmatrix}f(αa,αb,αc,αd)&g(αa,αb,αc,αd)&h(αa,αb,αc,αd)\\0&f(αa,αb,αc,αd)&i(αa,αb,αc,αd)\\0&0&f(αa,αb,αc,αd)\end{bmatrix}$
The right-hand side will then be
$αd(A)=\begin{bmatrix}αf(a,b,c,d)&αg(a,b,c,d)&αh(a,b,c,d)\\0&αf(a,b,c,d)&αi(a,b,c,d)\\0&0&αf(a,b,c,d)\end{bmatrix}$
Thus,
$f(αa,αb,αc,αd)=αf(a,b,c,d)$ $g(αa,αb,αc,αd)=αg(a,b,c,d)$ $h(αa,αb,αc,αd)=αh(a,b,c,d)$ $i(αa,αb,αc,αd)=αi(a,b,c,d)$
Therfore f,g,h,i are linear transformations
(3) Now to show how condition (3) holds. We will be using matrix A and matrix B from above. As before, we will start with proving the left side of the equal sign, and then progress to proving the right side, therefore showing that condition (3) is true as well.
$A=\begin{bmatrix}(a&b&c\\0&a&d\\0&0&a\end{bmatrix}$ $B=\begin{bmatrix}x&y&z\\0&x&v\\0&0&x\end{bmatrix}$
$d(AB)=Ad(B)+d(A)B$ $\to$
$\begin{bmatrix}f(ax,ay+bx,az+bv+cx,av+dx)&g(ax,ay+bx,az+bv+cx,av+dx)&h(ax,ay+bx,az+bv+cx,av+dx)\\0&f(ax,ay+bx,az+bv+cx,av+dx)&i(ax,ay+bx,az+bv+cx,av+dx)\\0&0&f(ax,ay+bx,az+bv+cx,av+dx)))\end{bmatrix}$
=$\begin{bmatrix}(af(x,y,z,v)+f(a,b,c,d)x &ag(x,y,z,v)+bf(x,y,z,v)+f(a,b,c,d)y+g(a,b,c,d)x&ah(x,y,z,v)+bi(x,y,z,v)+cf(x,y,z,v)+f(a,b,c,d)z+g(a,b,c,d)v+h(a,b,c,d)\\0&af(x,y,z,v)+f(a,b,c,d)x&ai(x,y,z,v)+df(x,y,z,v)+f(a,b,c,d)v+i(a,b,c,d)x\\0&0&0)\end{bmatrix}$
Since the above matrices equal each other, we can say that corresponding entries are equal to each other as well. From this point, we will match up each corresponding element in the matrices and set them equal to each other, formulating a set of 4 equations.
$Equation 1: f(ax,ay+bx,az+bv+cx,av+dx)= af(x,y,z,v)+f(a,b,c,d)x$ $Equation 2: g(ax,ay+bx,az+bv+cx,av+dx)=$ $ag(x,y,z,v)+bf(x,y,z,v)+f(a,b,c,d)y+g(a,b,c,d)x$ $Equation3:h(ax,ay+bx,az+bv+cx,av+dx)=ah(x,y,z,v)+bi(x,y,z,v)+cf(x,y,z,v)+f(a,b,c,d)z+g(a,b,c,d)v+h(a,b,c,d)x$ $Equation4:i(ax,ay+bx,az+bv+cx,av+dx)=ai(x,y,z,v)+df(x,y,z,v)+f(a,b,c,d)v+i(a,b,c,d)x$
Now that we have equations to work with, we would like to then simplify these equations so that we may then be able solve them. We will start with equation 1.
$Equation 1: f(ax,ay+bx,az+bv+cx,av+dx)= af(x,y,z,v)+f(a,b,c,d)x$
Starting with the left hand side, $f(ax,ay+bx,az+bv+cx,av+dx)$ $=f[(ax,0,0,0)+(0,ay+bx,0,0)+(0,0,az+bv+cx,0)+(0,0,0,av+dx)]$ $=axf(1,0,0,0)+ay+bx(0,1,0,0)+az+bv+cx(0,0,1,0)+av+dx(0,0,0,1)]$
If we assign $f(1,0,0,0)=a_1, f(0,1,0,0)=a_2, f(0,0,1,0)=a_3, f(0,0,0,1)=a_4$, we then have the following equation: $axa_1+aya_2+bxa_2+ a_(2 ) a_3+bva_3+ cxa_3+ava_4+dxa_4$
Following the same process with the rest of the equation givs:
$axa_1+aya_2+bxa_2+ aza_3+bva_3+ cxa_3+ava_4+dxa_4= axa_1+aya_2+aza_3+ ava_4+ aa_1 x+ba_2 x+ca_3 x+ da_4 x$
When we cancel out terms on either side, we get $bva_3=aa_1 x$.
If we remember,all $a,b,c,d,x,y,z,v Є R$, we can substitute and solve to further simplify the equation. If we set $a=0$, we then get that $a_1=a_3=0$ Taking our equation $f(a,b,c,d)=aa_1+ba_2+ca_3+da_4$, and substituting in our zero values, we are left with $f(a,b,c,d)=ba_2+da_4$. We can use our new information to simplify the other 3 equations. Let us begin with equation 2.
Equation 2: Using the same process as above, we will begin to simplify equation 2: $g(ax,ay+bx,az+bv+cx,av+dx)=$ $ag(x,y,z,v)+bf(x,y,z,v)+f(a,b,c,d)y+g(a,b,c,d)x$
If we assign $g(1,0,0,0)=b_1, g(0,1,0,0)=b_2, g(0,0,1,0)=b_3, g(0,0,0,1)=b_4$,
we get $axb_1+ayb_2+bxb_2+azb_3+bvb_3+cxb_3+avb_4+dxb_4= axb_1+ayb_2+azb_3+avb_4+ bxa_1+bya_2+bza_3+bva_4+ aya_1+bya_2+cya_3+dya_4+ axb_1+bxb_2+cxb_3+dxb_4$
Canceling out terms on either side and also canceling out any terms that have $a_1$ and $a_3$ since they equal 0, we get:
$bvb_3=bya_2+bva_4+bya_2+dya_4+axb_1$
If we have $b=0$ and $a=0$, then we are left with $0=dya_4$, therefore $a_4=0$ $bvb_3=bya_2+bva_4+bya_2+dya_4+axb_1$
Taking the same equation and setting $b=0$ and $v=0$, we then have $0=axb_1$, therefore $b_1=0$
$bvb_3=bya_2+bva_4+bya_2+dya_4+axb_1$
Once again taking the same equation, this time setting $b=1$ and $v=0$, gives us $0=2ya_2$, therefore, $a_2=0$
$bvb_3=bya_2+bva_4+bya_2+dya_4+axb_1$ and once again taking the same equation, this time just canceling out all the terms that are equal to 0, gives us $0=bvb_3$, which means that $b_3=0$ Giving us $g(a,b,c,d)=ab_1+bb_2+cb_3+db_4=bb_2+db_4$
So far, we have solved and gotten that $a_1=a_2=a_3=a_4=0$ and $b_1=b_2=0$ . If we apply this to equation 1: $f(a,b,c,d)=ba_2+da_4$, we now have $f(a,b,c,d)=0$
Equation 3: $h(ax,ay+bx,az+bv+cx,av+dx)=ah(x,y,z,v)+bi(x,y,z,v)+cf(x,y,z,v)+f(a,b,c,d)z+g(a,b,c,d)v+h(a,b,c,d)x$
Using the same process as above and assigning $h(1,0,0,0)=c_1, h(0,1,0,0)=c_2, h(0,0,1,0)=c_3, h(0,0,0,1)=c_4$, and if we assign $i(1,0,0,0)=d_1, i(0,1,0,0)=d_2, i(0,0,1,0)=d_3, i(0,0,0,1)=d_4$ we then have the following equation:
Equation 3 is now $axc_1+ayc_2+bxc_2+azc_3+bvc_3+cxc_3+avc_4+dxc_4= axc_1+ayc_2+azc_3+avc_4+ bxd_1+byd_2+bzd_3+bvd_4+ cxa_1+cya_2+cza_3+cva_4+aza_1+bza_2+cza_3+dza_4+ avb_1+avb_2+cvb_3+dvb_4+ axc_1+bxc_2+cxc_3+dxc_4$
Canceling out all terms gives $bvc_3= bxd_1+byd_2+bzd_3+bvd_4+avb_2+axc_1+dvb_4$ If we assign $b=0$ and $a=0$, we then have $0=dvb_4$, so therefore $b_4=0$
Taking the same equation $bvc_3= bxd_1+byd_2+bzd_3+bvd_4+avb_2+axc_1+dvb_4$ and now setting $b=0$ and $v=0$, we have $c_1=0$
$bvc_3= bxd_1+byd_2+bzd_3+bvd_4+avb_2+axc_1+dvb_4$ Taking this equation once again, and deleting the last two terms since we now know that they are 0, gives us $bvc_3= bxd_1+byd_2+bzd_3+bvd_4+avb_2$ and if we set $b=0$, we now get $0=avb_2$, which tells us that$ b_2=0$
Now, taking Equation 2 from above $g(a,b,c,d)=ab_1+bb_2+cb_3+db_4=bb_2+db_4$, we now know that $g(a,b,c,d)=0$
$h(a,b,c,d)=ac_1+bc_2+cc_3+dc_4=bc_2+cc_3+dc_4$ Equation 4: $i(ax,ay+bx,az+bv+cx,av+dx)=ai(x,y,z,v)+df(x,y,z,v)+f(a,b,c,d)v+i(a,b,c,d)x$ . Using the same process as above to simplify Equation 4, we get:
$axd_1+ayd_2+bxd_2+azd_3+bvd_3+avd_4+dxd_4=axd_1+ayd_2+azd_3+avd_4+dxa_1+dya_2+dza_3+dva_4+dxa_1+dya_2+dza_3+dva_4+ava_1+bva_2+cva_3+dva_4 axd_1+bxd_2+cxd_3+dxd_4$
Canceling out all terms gives $bvd_3=axd_1+cxd_3$ If we set $x=0$, we get $d_3=$0 And if $d_3=0$, then $d_1=0$ $i(a,b,c,d)=ad_1+bd_2+cd_3+dd_4=bd_2+ dd_4$
In my conclusion, I have that
$f(a,b,c,d)=0$
$g(a,b,c,d=0$
$h(a,b,c,d)=bc_2+cc_3+dc_4$
$i(a,b,c,d)=dd_4$
Pardon me for elaborating on loup blanc’s solution for cele’s benefit here; my elaboration would simply be too long for a comment.
For any $ A,B,C,D \in \Bbb{R} $, we have $$ \begin{bmatrix} A & B & C \\ 0 & A & D \\ 0 & 0 & A \end{bmatrix} = A \underbrace {\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}}_{I_{3}} + B \underbrace {\begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}}_{E_{12}} + C \underbrace {\begin{bmatrix} 0 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}}_{E_{13}} + D \underbrace {\begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix}}_{E_{23}}. $$ The matrices $ I_{3} $, $ E_{12} $, $ E_{13} $ and $ E_{23} $ thus form a basis for the $ \Bbb{R} $-vector space $ {\text{DT}_{3}}(\Bbb{R}) $. Note also that $ {\text{DT}_{3}}(\Bbb{R}) $ is an $ \Bbb{R} $-algebra under matrix multiplication.
Conditions $ (1) $, $ (2) $ and $ (3) $ say that the map $ d: {\text{DT}_{3}}(\Bbb{R}) \to {\text{DT}_{3}}(\Bbb{R}) $ is a derivation, i.e., a linear transformation that obeys the Leibniz Rule.
We know from linear algebra that a linear transformation is entirely determined by its values on a vector-space basis. Hence, $ d $, being a linear transformation, is entirely determined by its values on $ I_{3} $, $ E_{12} $, $ E_{13} $ and $ E_{23} $.
In what follows, $ \bullet $ shall denote matrix multiplication.
Firstly, $$ d(I_{3}) = d(I_{3} \bullet I_{3}) = I_{3} \bullet d(I_{3}) + d(I_{3}) \bullet I_{3} = 2 \cdot d(I_{3}), $$ which gives us $ d(I_{3}) = \mathbf{0}_{3} $.
Secondly, $$ d(E_{12} \bullet E_{12}) = E_{12} \bullet d(E_{12}) + d(E_{12}) \bullet E_{12}. $$ However, $ E_{12} \bullet E_{12} = \mathbf{0}_{3} $, so $ d(E_{12} \bullet E_{12}) = \mathbf{0}_{3} $, which yields $$ (\clubsuit) \qquad E_{12} \bullet d(E_{12}) = - d(E_{12}) \bullet E_{12}. $$ Suppose that $ d(E_{12}) = \begin{bmatrix} A & B & C \\ 0 & A & D \\ 0 & 0 & A \end{bmatrix} $. Then $ (\clubsuit) $ implies that \begin{align} \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} \begin{bmatrix} A & B & C \\ 0 & A & D \\ 0 & 0 & A \end{bmatrix} & = - \begin{bmatrix} A & B & C \\ 0 & A & D \\ 0 & 0 & A \end{bmatrix} \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}, \\ \begin{bmatrix} 0 & A & D \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} & = - \begin{bmatrix} 0 & A & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}. \end{align} Hence, $ D = 0 $ and $ A = - A $, which also gives us $ A = 0 $. Therefore, $$ d(E_{12}) = \begin{bmatrix} 0 & B & C \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}. $$
Thirdly, $$ d(E_{13} \bullet E_{13}) = E_{13} \bullet d(E_{13}) + d(E_{13}) \bullet E_{13}. $$ However, $ E_{13} \bullet E_{13} = \mathbf{0}_{3} $, so $ d(E_{13} \bullet E_{13}) = \mathbf{0}_{3} $, which yields $$ (\spadesuit) \qquad E_{13} \bullet d(E_{13}) = - d(E_{13}) \bullet E_{13}. $$ Suppose that $ d(E_{13}) = \begin{bmatrix} A' & B' & C' \\ 0 & A' & D' \\ 0 & 0 & A' \end{bmatrix} $. Then $ (\spadesuit) $ implies that \begin{align} \begin{bmatrix} 0 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} \begin{bmatrix} A' & B' & C' \\ 0 & A' & D' \\ 0 & 0 & A' \end{bmatrix} & = - \begin{bmatrix} A' & B' & C' \\ 0 & A' & D' \\ 0 & 0 & A' \end{bmatrix} \begin{bmatrix} 0 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}, \\ \begin{bmatrix} 0 & 0 & A' \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} & = - \begin{bmatrix} 0 & 0 & A' \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}. \end{align} Hence, $ A' = - A' $, which gives us $ A' = 0 $. Therefore, $$ d(E_{13}) = \begin{bmatrix} 0 & B' & C' \\ 0 & 0 & D' \\ 0 & 0 & 0 \end{bmatrix}. $$
Fourthly, $$ d(E_{23} \bullet E_{23}) = E_{23} \bullet d(E_{23}) + d(E_{23}) \bullet E_{23}. $$ However, $ E_{23} \bullet E_{23} = \mathbf{0}_{3} $, so $ d(E_{23} \bullet E_{23}) = \mathbf{0}_{3} $, which yields $$ (\heartsuit) \qquad E_{23} \bullet d(E_{23}) = - d(E_{23}) \bullet E_{23}. $$ Suppose that $ d(E_{23}) = \begin{bmatrix} A'' & B'' & C'' \\ 0 & A'' & D'' \\ 0 & 0 & A'' \end{bmatrix} $. Then $ (\heartsuit) $ implies that \begin{align} \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix} \begin{bmatrix} A'' & B'' & C'' \\ 0 & A'' & D'' \\ 0 & 0 & A'' \end{bmatrix} & = - \begin{bmatrix} A'' & B'' & C'' \\ 0 & A'' & D'' \\ 0 & 0 & A'' \end{bmatrix} \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix}, \\ \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & A'' \\ 0 & 0 & 0 \end{bmatrix} & = - \begin{bmatrix} 0 & 0 & B'' \\ 0 & 0 & A'' \\ 0 & 0 & 0 \end{bmatrix}. \end{align} Hence, $ B'' = 0 $ and $ A'' = - A'' $, which also gives us $ A'' = 0 $. Therefore, $$ d(E_{23}) = \begin{bmatrix} 0 & 0 & C'' \\ 0 & 0 & D'' \\ 0 & 0 & 0 \end{bmatrix}. $$
Organizing the information that we have, we find that there exist $ p,q,r,s,t,u,v \in \Bbb{R} $ such that \begin{align} d(E_{12}) & = \begin{bmatrix} 0 & p & q \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix},\\ d(E_{13}) & = \begin{bmatrix} 0 & r & s \\ 0 & 0 & t \\ 0 & 0 & 0 \end{bmatrix},\\ d(E_{23}) & = \begin{bmatrix} 0 & 0 & u \\ 0 & 0 & v \\ 0 & 0 & 0 \end{bmatrix}. \end{align} We proceed to show that $ r = t = 0 $ and $ s = p + v $.
Observe that $$ d((E_{12} + E_{23}) \bullet (E_{12} + E_{23})) = (E_{12} + E_{23}) \bullet d(E_{12} + E_{23}) + d(E_{12} + E_{23}) \bullet (E_{12} + E_{23}). $$ However, $ (E_{12} + E_{23}) \bullet (E_{12} + E_{23}) = E_{13} $, so $$ d(E_{13}) = (E_{12} + E_{23}) \bullet d(E_{12} + E_{23}) + d(E_{12} + E_{23}) \bullet (E_{12} + E_{23}). $$ This implies that \begin{align} \begin{bmatrix} 0 & r & s \\ 0 & 0 & t \\ 0 & 0 & 0 \end{bmatrix} & = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix} \begin{bmatrix} 0 & p & q + u \\ 0 & 0 & v \\ 0 & 0 & 0 \end{bmatrix} + \begin{bmatrix} 0 & p & q + u \\ 0 & 0 & v \\ 0 & 0 & 0 \end{bmatrix} \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix} \\ & = \begin{bmatrix} 0 & 0 & v \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} + \begin{bmatrix} 0 & 0 & p \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} \\ & = \begin{bmatrix} 0 & 0 & p + v \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}. \end{align} Hence, $ r = t = 0 $ and $ s = p + v $.
We now have a more complete description of $ d $, namely, there exist $ p,q,u,v \in \Bbb{R} $ such that \begin{align} d(E_{12}) & = \begin{bmatrix} 0 & p & q \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix},\\ d(E_{13}) & = \begin{bmatrix} 0 & 0 & p + v \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}, \\ d(E_{23}) & = \begin{bmatrix} 0 & 0 & u \\ 0 & 0 & v \\ 0 & 0 & 0 \end{bmatrix}. \end{align}
Finally, to show that $ p $, $ q $, $ u $ and $ v $ have no more interdependence, consider $$ M = \begin{bmatrix} A & B & C \\ 0 & A & D \\ 0 & 0 & A \end{bmatrix} \quad \text{and} \quad M' = \begin{bmatrix} A' & B' & C' \\ 0 & A' & D' \\ 0 & 0 & A' \end{bmatrix}. $$ Then \begin{align} d(M) & = A \cdot d(I_{3}) + B \cdot d(E_{12}) + C \cdot d(E_{13}) + D \cdot d(E_{23})\\ & = A \cdot \mathbf{0}_{3} + B \begin{bmatrix} 0 & p & q \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} + C \begin{bmatrix} 0 & 0 & p + v \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} + D \begin{bmatrix} 0 & 0 & u \\ 0 & 0 & v \\ 0 & 0 & 0 \end{bmatrix} \\ & = \begin{bmatrix} 0 & B p & B q + C (p + v) + D u \\ 0 & 0 & D v \\ 0 & 0 & 0 \end{bmatrix}. \end{align} Similarly, $ d(M') = \begin{bmatrix} 0 & B' p & B' q + C' (p + v) + D' u \\ 0 & 0 & D' v \\ 0 & 0 & 0 \end{bmatrix} $. Next, observe that \begin{align} M \bullet d(M') & = \begin{bmatrix} 0 & A B' p & A B' q + A C' (p + v) + A D' u + B D' v \\ 0 & 0 & A D' v \\ 0 & 0 & 0 \end{bmatrix}, \\ d(M) \bullet M' & = \begin{bmatrix} 0 & A' B p & B D' p + A' B q + A' C (p + v) + A' D u \\ 0 & 0 & A' D v \\ 0 & 0 & 0 \end{bmatrix}. \end{align} Hence, $$ M \bullet d(M') + d(M) \bullet M' = \\ \begin{bmatrix} 0 & (A B' + A' B) p & (A B' + A' B) q + (A C' + A' C + B D') (p + v) + (A D' + A' D) u \\ 0 & 0 & (A D' + A' D) v \\ 0 & 0 & 0 \end{bmatrix}. $$ As $$ M \bullet M' = \begin{bmatrix} A A' & A B' + A' B & A C' + A' C + B D' \\ 0 & A A' & A D' + A' D \\ 0 & 0 & A A' \end{bmatrix}, $$ we obtain $$ d(M \bullet M') = \\ \begin{bmatrix} 0 & (A B' + A' B) p & (A B' + A' B) q + (A C' + A' C + B D') (p + v) + (A D' + A' D) u \\ 0 & 0 & (A D' + A' D) v \\ 0 & 0 & 0 \end{bmatrix} $$ as well. Therefore, $ d(M \bullet M') = M \bullet d(M') + d(M) \bullet M' $ without any further hypotheses.
Conclusion: A map $ d: {\text{DT}_{3}}(\Bbb{R}) \to {\text{DT}_{3}}(\Bbb{R}) $ is a derivation if and only if there exist scalars $ p $, $ q $, $ u $ and $ v $ such that $$ d \! \left( \begin{bmatrix} A & B & C \\ 0 & A & D \\ 0 & 0 & A \end{bmatrix} \right) = \begin{bmatrix} 0 & B p & B q + C (p + v) + D u \\ 0 & 0 & D v \\ 0 & 0 & 0 \end{bmatrix}. $$