Past exam of the options II course of the University of Oxford / 2012 / F / F2 / c / i / Solution

The chain rule givesIts determinant is $-1$, so it is invertible. The map $G$ is smooth and $G(0,0)=(0,0)$. The inverse function theorem therefore yields an open neighbourhood $U$ of the origin for which $G(U)$ is open, contains the origin, and