Past exam of the mathematics course of the University of Cambridge / 2019 / ib / Paper 2 / 3E (Analysis II)

For $x\ne 0$, all component functions are continuously differentiable and the Jacobian matrix isAt a point $(0,y_0)$, the first component cannot be differentiable: along the first coordinate axis its difference quotient contains $h^{1/3}/h=h^{-2/3}$, which is unbounded as $h\to 0$. Therefore $f$ is continuously differentiable exactly at the points with $x\ne 0$.

At a point with $xy\ne 0$, the determinant of the derivative isThe inverse function theorem therefore supplies neighbourhoods on which $f$ has a continuously differentiable, hence differentiable, inverse. Concretely, if $(u,v)=f(x,y)$, the local inverse is the restriction ofwhich is continuously differentiable near the image whenever $y\ne 0$.