jacobian matrix of $f^{-1}$ at $(1,0,2)$ given $f(x,y,z)$

Question

If $f(x,y,z)=(sin(xyz),(x+x^2)*cos(y),y)$ and $f$ has a local inverse in the neighborhood of $(0,1,1)$, how do I find the jacobian matrix of this inverse at $(1,0,2)$? I know from definition that $J_{f^{-1}}=J_f(f^{-1}(y))^{-1}$ but don't really know how to proceed from here.

Answer 1

Use the inverse function theorem; (f^-1(x))'=1/f(x)' that is, derivative of the inverse function is the reciprocal of derivative of the function in the neighbourhood where f has a local inverse.

Using that, you can calculate the Jacobian of the inverse function.