"Let $h=f\circ g$, where $f:\mathbb{R}^2 \rightarrow \mathbb{R}$ and $g:\mathbb{R}^2 \rightarrow \mathbb{R}^2$ is defined by $g(x,y)=(2x^2y,3y-2x)$. Find the derivative $h'(-2,-3), h_x(-2,-3), h_y(-2,-3)$ if $f_x(-24,-5)=3$ and $f_y(-24,-5)=3$."
I tried to draw a diagram, but I got confused a little bit because of so many variables. I then tried to find the answer by guessing an $f$ to deduce the solution from it but for different $f$'s, I found different answers. (I took $f(x,y)=3x+3y$ and $f(x,y)=x^2+y^2+51x+51y$).
Also, I am not sure what $h'(-2, -3)$ is supposed to mean. When we talk about the derivative of a multivariable function, I thought we meant the partial derivatives. If the question is talking about the gradient, then isn't asking for $h_x$ and $h_y$ redundant?
Accordingly Rudin W. - Principles of mathematical analysis-(1976), p211-213 if we have open set $E \subset \mathbb{R^n}$ and $f$ mapps it to $\mathbb{R^m}$, then for $x \in E$ we define total derivative as linear transformation $A$ of $\mathbb{R^n}$ into $\mathbb{R^m}$ such that $$\lim\limits_{h \to 0}\frac{|f(x+h)-f(x)-Ah|}{|h|}=0$$ then we call $f$ differentiable and write $f'(x)=A$. $A$ is so called Jacobian matrix and is calculated using partial derivatives. In case $m=1$ we have one row in matrix, which is the transpose of the gradient.
If denote $(u,v)=g(x,y) = (2x^2y,3y-2x) $, then we have: $$\frac{\partial h}{\partial x}=\frac{\partial f}{\partial u}\frac{\partial u}{\partial x}+\frac{\partial f}{\partial v}\frac{\partial v}{\partial x}=\frac{\partial f}{\partial u}4xy-\frac{\partial f}{\partial v}2$$
So, hope, calculate $h', h_x,h_y$ shouldn't be problem.