Differentiation of one function with respect to another in multivariable calculus?

There is no such thing as this kind of derivative, without more information. Everyone is being imprecise when they talk about the derivative of $g(x, y) = x^2 + y^2$ with respect to $x$: this is actually not enough information. What they mean is the derivative with respect to $x$, holding $y$ fixed; this second clause is left implicit but it's actually crucial.

Here's a simpler example: what's the derivative of $g(x, y) = x^2 + y^2$ with respect to $a = x + y$? The answer is that there's no way of telling what this means until you tell me what is being left fixed; for example, if I'm fixing $y$ then we can rewrite

$$g = (a - y)^2 + y^2, \frac{\partial g}{\partial a} = 2(a - y) = 2x$$

but if I'm fixing $b = x - y$ then we can rewrite

$$g = \left( \frac{a+b}{2} \right)^2 + \left( \frac{a-b}{2} \right)^2, \frac{\partial g}{\partial a} = \frac{a+b}{2} + \frac{a-b}{2} = a = x + y$$

and this is not the same!

Formally, what is needed is to specify a vector field. On $\mathbb{R}^n$ we can specify a vector field by choosing coordinate functions $x_1, \dots x_n : \mathbb{R}^n \to \mathbb{R}$, which specifies a dual basis $\frac{\partial}{\partial x_1}, \dots \frac{\partial}{\partial x_n}$ of vector fields. Crucial here is that despite what the notation says, the construction of $\frac{\partial}{\partial x_i}$ depends on the entire rest of the $x_j$: it's "the derivative with respect to $x_i$, keeping the other $x_j$ fixed." If you just have some fixed coordinate function $x$, what you can define is the differential $1$-form $dx$, but you don't have a way of turning this $1$-form into a vector field.