Let $f(u,v) = c$ where $u(x,y) , v(x,y)$ are functions and $c$ is constant. Can we conclude $\frac{\partial f}{\partial v} = \frac{\partial f}{\partial u} = 0$ ? It really sounds confusing to me but I've tried many examples and also the definition of partial derivative , and it was true ! What's the problem here ?
Main question : Suppose $f$ is a differentiable function . If $z$ is a differentiable function with respect to $x$ and $y$ and defined in $f(xz,yz) = 1$ prove that : $x\frac{\partial z}{\partial x} + y\frac{\partial z}{\partial y} = -z$
There is some confusion being caused by the employment of dummy variables. Strictly speaking, if we have a differentiable function $f\colon \mathbf R^2\to\mathbf R$, then we can write it as $f = f(x,y) = f(u,v) = f(\uparrow,\downarrow), \dots$. The partial derivatives of $f$ with respect to the dummy variables we are using for $\mathbf R^2$ then use the same dummy variables by convention.
Now, in the question you are trying to solve, you are asked to show that for all $x,y$ such that $$ f\big(x\cdot z(x,y), y\cdot z(x,y)\big) = 1, $$ the equation $$ -z(x,y) = x\frac{\partial z}{\partial x} + y\frac{\partial z}{\partial y} $$ holds. We are not saying that $f = f(\text{dummy variable 1},\text{dummy variable 2})$ is the constant function $1$, in which case we of course would have $\partial_1f = \partial_2 f = 0$. Instead, we are saying, look at all the points $(x,y)\in\mathbf R^2$ at which $$ f\big(x\cdot z(x,y), y\cdot z(x,y)\big) = 1\quad\text{holds}. $$ For any such point $(x,y)$, show that $$ -z(x,y) = x\frac{\partial z}{\partial x} + y\frac{\partial z}{\partial y},\quad\text{also holds.} $$