Can someone give a proof or point to a proof of the following.
For $y = h(x) = y(x)$ where $h: \mathbb{R}^n \rightarrow \mathbb{R}^n$ is a differentiable function with a differential inverse $x = h^{-1}(y) = x(y)$. (i.e., $h$ is a coordinate transformation.) Let, $$ J_y(x) = \frac{\partial h(x)}{\partial x} \in \mathbb{R}^{n\times n} $$ Then $J_{y}(x)$ is non-singular and $J_{x}(y) = J^{-1}_{y}(x)$
I am not able to prove that $J_{y}(x)$ is non-singular. Assuming that it is non-singular though I can prove the second part using the chain rule. For $x = h^{-1}(y) = h(h(x))$
$$ \begin{align} \frac{\partial x}{\partial x} &= \frac{\partial h^{-1}(y(x))}{\partial x} \\ I &= \frac{\partial h^{-1}(y)}{\partial y} \frac{\partial y(x)}{\partial x} \\ I &= J_{x}(y)J_{y}(x) \end{align} $$
Your computation doesn't assume that $J_y$ is nonsingular, it simply shows that $I=J_x J_y$. Then $1=\det I=\det(J_x)\det(J_y)$, so $\det(J_y)$ can't be zero.