Show that the function $z = z(x, y)$ determined implicitly by the equation $F(x - az, y - bz) = 0$, in which $F$ is a differentiable function, satisfies the equation
$$ a\frac{\partial z}{\partial x} + b\frac{\partial z}{\partial y} = 1 $$
My idea: define $c = x - az$ and $d = y - bz$. Then we get $F(c, d)$. Apply the chain rule there. However, this leads to
$$ 0 = \frac{\partial F}{\partial x} =\frac{\partial F}{\partial c}\frac{\partial c}{\partial z}\frac{\partial z}{\partial x} + \frac{\partial F}{\partial b}\frac{\partial b}{\partial z}\frac{\partial z}{\partial x} $$
And this leads to
$$ \frac{\partial z}{\partial x} = 0 $$
The same thing would happen to $ \frac{\partial z}{\partial y} $.
So the given equation isn't satisfied by this and I had no other ideas. The hard part is finding these partial derivatives, after that it's just a matter of verifying the equation.
Let $\quad \begin{cases} X=x-az &,& \frac{\partial X}{\partial x}=1-a \frac{\partial z}{\partial x} &,& \frac{\partial X}{\partial y}= -a\frac{\partial z}{\partial y}\\ Y=y-bz &,& \frac{\partial Y}{\partial x}= -b\frac{\partial z} {\partial x} &,& \frac{\partial Y}{\partial y}= 1-b\frac{\partial z}{\partial y} \end{cases}$
Differentiate $\quad F(X,Y)=0$ : $$\frac{\partial F}{\partial X}\frac{\partial X}{\partial x}dx +\frac{\partial F}{\partial X}\frac{\partial X}{\partial y}dy +\frac{\partial F}{\partial Y}\frac{\partial Y}{\partial x}dx +\frac{\partial F}{\partial Y}\frac{\partial Y}{\partial x}dy =0$$
$$\frac{\partial F}{\partial X}\left(1-a\frac{\partial z}{\partial x}\right)dx +\frac{\partial F}{\partial X}\left(-a\frac{\partial z}{\partial y}\right)dy +\frac{\partial F}{\partial Y}\left(-b\frac{\partial z} {\partial x}\right)dx +\frac{\partial F}{\partial Y}\left(1-b\frac{\partial z}{\partial y}\right)dy=0$$ This implies : $$\begin{cases} \frac{\partial F}{\partial X}\left(1-a\frac{\partial z}{\partial x}\right)+\frac{\partial F}{\partial Y}\left(-b\frac{\partial z} {\partial x}\right)=0 \\ \frac{\partial F}{\partial X}\left(-a\frac{\partial z}{\partial y}\right)+\frac{\partial F}{\partial Y}\left(1-b\frac{\partial z}{\partial y}\right)=0 \end{cases}$$ $$\left(1-a\frac{\partial z}{\partial x}\right)\left(1-b\frac{\partial z}{\partial y}\right)=\left(-b\frac{\partial z} {\partial x}\right)\left(-a\frac{\partial z}{\partial y}\right)$$ $$1-a\frac{\partial z}{\partial x}-b\frac{\partial z}{\partial y}=0$$ $$a\frac{\partial z}{\partial x}+b\frac{\partial z}{\partial y}=1$$
ADDITION after your comment :
$$\frac{\partial F}{\partial x} =\frac{\partial F}{\partial X}\left(\frac{\partial X}{\partial x}+\frac{\partial X}{\partial z}\frac{\partial z}{\partial x}\right)+\frac{\partial F}{\partial Y}\left(\frac{\partial Y}{\partial x}+\frac{\partial Y}{\partial z}\frac{\partial z}{\partial x}\right)$$
$$\frac{\partial F}{\partial y} =\frac{\partial F}{\partial X}\left(\frac{\partial X}{\partial y}+\frac{\partial X}{\partial z}\frac{\partial z}{\partial y}\right)+\frac{\partial F}{\partial Y}\left(\frac{\partial Y}{\partial y}+\frac{\partial Y}{\partial z}\frac{\partial z}{\partial y}\right)$$
Then you get the same equations than above.