For $f(x,y) = (x^3 - y^2, \sin{x} - \ln{y})$ f-inverse exists and is differentiable in a non-empty set around $(-1,0)$. Find $D(f^{-1})$ at $(-1,0)$.
Seemingly this is an Inverse Function Theorem problem. So I have my J matrix ($a_{ij} = \frac{\partial f_i}{\partial x_j}$):
$$\begin{vmatrix}{3x^2}&{-2y}\\{\cos{x}}&{\frac{-1}{y}}\\ \end{vmatrix}$$ and its determinant is $2y\cos{x} - \frac{3x^2}{y}$.
However, evaluating at $(-1,0)$ yields $\frac{-3}{0}$.
Can I use the IFT here where the determinant is undefined? Is it sufficient that it's not zero in the area around $(-1,0)$? Or did I make a mistake in calculation? How else could I answer this question?
You want the inverse to exist around $(-1,0)$, not $f$ to be invertible around $(-1,0)$ (as $f$ isn't defined at $(-1,0)$ this does not really make sense, as you observe). So we start to find $(x,y)$ such that $f(x,y) = (-1,0)$. We have $$ x^3 - y^2 = -1, \quad \sin x = \log y$$ as solution is $(0, 1)$. Now use the IFT to prove that $f$ is invertible around $(0,1)$. We have $$ \det Df(0,1) = 0 \cdot (-1) - (-2) \cdot 1 = 2 \ne 0. $$ For computing $D(f^{-1})(0,1)$ recall that $$ Df^{-1}(-1,0) = Df\bigl(f^{-1}(-1,0)\bigr)^{-1} = Df(0,1)^{-1} $$