I think this should be an easy question? However, a quick google search has not revealed the confirmation I want, so I am posting here.
If we have a function $f(x,y)$ and a monotonic transformation $g$. Let us now consider a monotonic transformation of $f$ using $g$: that is, let $v(x,y) = g(f(x,y))$. (Note that $g: I \to \mathbb{R}$, where $I$ is some interval on the reals.
I see no reason why $v$ would not be a composition of functions, so I believe we can say that $v(x,y) = (g \circ f) (x,y)$ My question is what is $\partial v \over \partial x$ (lets say evaluated at a point q = (a,b)).
What a formula be $\frac{\partial g}{\partial x} (f(q)) \frac{\partial f}{\partial x} (q)$? (where ($f(q)$) and $(q)$ mean evaluated at those points). Why or why not?
Yes you've got it: $v$ is a composition of functions, and the chain rule applies, just the way you wrote it (but change the partial derivative of $g$ to the usual single-variable derivative!). "Why" it's correct: because you're holding $y$ fixed while taking the derivative of:
$$x \mapsto g(f(x, y)) \colon \mathbb{R} \to \mathbb{R}$$
(Perhaps the domain of this function isn't all of $\mathbb{R}$, but anyway it's a subset thereof.)