So I recently encountered a problem in multivariable calculus, where one had to calculate $\frac{d(u,v)}{d(x,y)}$ for $$\left\{\begin{matrix} u = f(x,y) = h(g(x,y)) \\ v=g(x,y) \\ \end{matrix}\right.$$ where $g$ is a function of two variables whereas $h$ is a function of one variable.
Forming the Jacobian determinant we get:
$$ \begin{vmatrix} f_x(x,y)&f_y(x,y) \\ g_x(x,y)&g_y(x,y) \\ \end{vmatrix}$$ which after applying the chain rule on $f(x,y)$ yields us:
$$ \begin{vmatrix} h'(g(x,y))g_x(x,y)&h'(g(x,y))g_y(x,y) \\ g_x(x,y)&g_y(x,y) \\ \end{vmatrix}$$
which then trivially results in that the determinant is $0$.
From what I've learned so far in linear algebra, the determinant presents a scalar factor during a linear transformation. By looking at our given function locally speaking, we can linearize it, for which we can find that the Jacobian matrix sort of act as our matrix in a linear transformation. With that in mind, the Jacobian determinant represents the scaling factor after applying the Jacobian matrix (locally speaking).
From Linear Algebra, I also know that when the determinant is $0$, we have that our transformation is non - invertible. Now my question is, in what way can we relate the choice of variables above, to what actually happens geometrically speaking? Personally, I'm thinking that since we are relating $u$ with $v$ through a function $h$, we are esentially eliminating our $x's$ and $y's$, hence that's why we get a determinant of 0.
Thank you for any ideas / thoughts that may help me further in understanding this.