I need to compute the commutator subgroup of Unitriangular matrix $UT_{3}(p)$.
My attempt:-
Since $UT_{3}(p) / Z(UT_{3}(p)) \simeq C_{p^2}$ and we know that the commutator subgroup is the smallest normal subgroup with abelian quotient hence the commutator subgroup should be $Z(UT_{3}(p))$.
This can be done more generally. For simplicity, we denote by $E_{12}$ the element
$$ \begin{bmatrix} 1 & 1 & & & \\ & 1 & & & \\ & & 1& & \\ & & & \ddots & \\ & & & & 1\\ \end{bmatrix} $$ here the other entries are $0$. In general, then let $E_{ij}$ denote the matrix obtained by inserting $1$ in the $(i,j)$-th place of the identity matrix, and with $i<j$ (so insertion is in upper part).
These matrices $E_{12}, E_{13}, \cdots, E_{1n}, E_{23},\cdots , E_{2n}, \cdots$ satisfy the following commutator relation $$[E_{ij}, E_{jk}]= E_{ik} \mbox{ and } [E_{ij}, E_{kl}]=1 \mbox{ for $j\neq k$}.$$ (see Alperin-Bell, Groups and Representations, Chapter on General Linear Groups)
Thus,we have $[E_{12}, E_{23}]=E_{13}$, $[E_{12}, E_{24}]=E_{14}$, and so on.
You may try to write for size-$4$ matrices. Then we see that the elements $$ \begin{bmatrix} 1 & 0& 1 & \\ & 1 & 0 & \\ & & 1 & 0\\ & & & 1 \end{bmatrix}, \begin{bmatrix} 1 & 0 & &1 \\ & 1 & 0 & \\ & & 1 & 0 \\ & & & 1 \end{bmatrix}, \begin{bmatrix} 1 & 0 & & \\ & 1 &0 & 1\\ & & 1 &0 \\ & & & 1 \end{bmatrix},$$ are inside the commutator subgroup, and they span the subgroup $$ N=\begin{Bmatrix} \begin{bmatrix} 1 & 0 & * & * \\ & 1 &0 & *\\ & & 1 &0 \\ & & & 1 \end{bmatrix}\colon * \in Z_p.\end{Bmatrix}.$$ Moreover $U(4,p)/N$ is abelian, where $N$ is generated by some commutators, by the property of commutator subgroup you mentioned, it follows $N$ is the comutator subgroup.