What is the derivative of $f(x)$ with respect to $g(x,y)$ 2 Suppose that we have a function $f(x)$ of one variable, like $f(x)=x$.

And I want to derive this function with respect to another multivariate function $g(x,y)=x-y$.

My hypothesis is that the derivative is equal to $0$:

$$\frac{{\partial f(x)}}{{\partial g(x,y)}}=0$$

My reason to think that is that $f(x)$ can never be expressed in terms of $g(x,y)$, so $g(x,y)$ does not impact directly $f(x)$ in any way. Also $x$ and $y$ are independent, meaning they are not related, there is no function between them.

Or maybe is undetermined?

But if that is true, I would need to prove it. I would appreciate if you could help me.
Basically it's a normal derivative, but the fact that it does operate on the variable $(x-y)$.

For example: 

$$\frac{\partial (x-y)^2}{\partial (x-y)}$$

By calling $(x-y) = z$,  you can find the usual derivative:

$$\frac{\text{d} z^2}{\text{d}z} = 2z \longrightarrow 2(x-y)$$

as it shall be.

In your case your function is simply $x$.

By a change of variable like $x-y = z$ you find that $x = z-y$ hence your derivative is like

$$\frac{\text{d} (z-y)}{\text{d}z} = \frac{\text{d}}{\text{d}z}(z-y) = 1$$

In any case, let me tell you that that is a very ill notation.
Basically it's a normal derivative, but the fact that it does operate on the variable $(x-y)$.

For example: 

$$\frac{\partial (x-y)^2}{\partial (x-y)}$$

By calling $(x-y) = z$,  you can find the usual derivative:

$$\frac{\text{d} z^2}{\text{d}z} = 2z \longrightarrow 2(x-y)$$

as it shall be.

In your case your function is simply $x$.

By a change of variable like $x-y = z$ you find that $x = z-y$ hence your derivative is like

$$\frac{\text{d} (z-y)}{\text{d}z} = \frac{\text{d}}{\text{d}z}(z-y) = 1$$

In any case, let me tell you that that is a very ill notation.