Let us suppose we are given a function $f(x,y)$, which can also be expressed as $g(s,t)$. If we are given either a pair of $x=x(s,t)$ and $y=y(s,t)$, or $s=s(x,y)$ and $t=t(x,t)$, is it possible to use the chain rule to compute the partial derivatives of $g$ in terms of partial derivatives of $f$?
In other words, given that $f(x,y)$ can be expressed as $h(s,t)$, are the following formulas correct?
$$\frac{\partial h}{\partial s}=\frac{\partial f}{\partial x}\frac{\partial x} {\partial s}+\frac{\partial f}{\partial y}\frac{\partial y}{\partial s}$$
$$\frac{\partial h}{\partial t}=\frac{\partial f}{\partial x}\frac{\partial x}{\partial t}+\frac{\partial f}{\partial y}\frac{\partial y}{\partial t}$$
If you want $f(x,y)=f(x(s,t),y(s,t))$ but thinking to $g$ as change of coordinates $g:\mathbb R^2\cdot\!\!\dashrightarrow\mathbb R^2$, then, for a scalar function $f:\mathbb R^2\cdot\!\!\dashrightarrow\mathbb R$, you can take the composition $$\mathbb R^2\cdot\!\!\stackrel g\dashrightarrow\mathbb R^2\cdot\!\!\stackrel f\dashrightarrow\mathbb R.$$ So, the Chain's Rule applied to $f\circ g$ gives $(f\circ g)'=f'(g)\cdot g'$ and this would be $$\left[\frac{\partial(f\circ g)}{\partial s}, \frac{\partial(f\circ g)}{\partial t}\right]=\left[\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}\right] \left[ \begin{array}{cc} \dfrac{\partial x}{\partial s}&\dfrac{\partial x}{\partial t}\\ \\ \dfrac{\partial y}{\partial s}&\dfrac{\partial y}{\partial t}\end{array} \right]. $$ That is $$\frac{\partial (f\circ g)}{\partial s}=\frac{\partial f}{\partial x}\frac{\partial x}{\partial s}+\frac{\partial f}{\partial y}\frac{\partial y}{\partial s},$$ and $$\frac{\partial (f\circ g)}{\partial t}=\frac{\partial f}{\partial x}\frac{\partial x}{\partial t}+\frac{\partial f}{\partial y}\frac{\partial y}{\partial t}.$$