Despite my question has a title similar to Partial Derivatives of $F(x,y,z)$ where $z = f(x,y)$ I didn't understand its solution. I'm sorry for my ignorance on this subject.
How do I calculate the z-partial derivative of $f(x, y, z)$ where z=x+y? Or in a general form, where $z=g(x, y)$? I'm asking because I don't know how to deal with, using a wrong notation for example, $\frac{\partial f(x, y) } {\partial (x+y)} $.
On
It seems there isn't a general procedure for it. As Thomas said, the way it is to calculate the derivative to z and, in the result, replace z by $z=g(x, y)$.
On
Remark: the notation $F_z(a,b,c)$ is a source of confusion. There is no $z$ in the expression; its a 'dummy argument' for the "third argument of the expression"; and after taking the derivation the arguments $a,b,c$ are then substituted in. So that would be $$F_z(a,b,c)=\left.\dfrac{\partial F(x,y,z)}{\partial z}\right\rvert_{(x,y,z)=(a,b,c)}$$
... I'd suggest using $F_1,F_2, F_3$ or such to indicate differentiation with respect to an argument, rather than a dummy variable.
using a wrong notation for example, $\frac{∂f(x,y)}{∂(x+y)}$
Yeap, so wrong. Use $\left.\frac{\partial f(w)}{\partial w}\right\rvert_{w=x+y}$ as below.
This is just $f'(x+y)$: the dervative of $f$ evaluated at $x+y$. Which is $f_1(x+y)$ in the suggested notation.
From the start $$F_1(a,b,c)(x-a)+F_2(a,b,c)(y-b)+F_3(a,b,c)(z-c)=0\tag 1$$
When we take $F(x,y,z)=z-f(x+y)$, we can evaluate these terms:
$$\begin{align}F_1(a,b,c) &=\dfrac{\partial[ c-f(a+b)]}{\partial a}\\&= -\dfrac{\partial f(a+b)}{\partial a}\\[1ex] &= -\left.\dfrac{\partial f(w)}{\partial w}\right\rvert_{w=a+b}\cdot\dfrac{\partial [a+b]}{\partial a}\\[1ex]&=-f_1(a+b)\\[2ex] F_2(a,b,c)&= -f_1(a+b)\\[2ex] F_3(a,b,c) &=\dfrac{\partial [c-f(a+b)]}{\partial c}\\&=1 \end{align}$$
So we obtain from (1) $$-f_1(x+y)~(x-a)- f_1(x+y)~(y-b)+ (z-c)=0\tag{2'}$$
When $f(x+y)=x+y$, that becomes: $$-(x-a)- (y-b)+ (z-c)=0\tag{2''}$$
The partial derivative notation is indicating differentiation with respect to a certain argument of a function depending on several variables - it does not take into account whether that argument has further dependencies. So if you have $$f(x,y,z) = z$$ then $$ \frac{\partial f}{\partial z} (x, y, g(x,y, t, \alpha))= 1 \text{ and } \frac{\partial f}{\partial x} (x, y, g(x,y, t, \alpha))= 0$$ regardless of what $g$ is.
As another example, if you look at $h(x,y, z) = z^2$ then
$$\frac{\partial h}{\partial z} = 2z $$ and, consequently, $$\frac{\partial h}{\partial z} (a, b, g(x,y)) = 2g(x,y) $$
If you want to differentiate, say, $f(x,y,x^2+y)$ and want to indicate you are interested in the dependency on $x$, you would write $$\frac{df}{dx}$$ and the result would be (in the example given above), according to the chain rule,
$$\frac{\partial f}{\partial x} +\frac{\partial f}{\partial y} \frac{dy}{dx} +\frac{\partial f}{\partial z} \frac{dz}{dx} = 0 + 0 + 2x $$