Past exam of the options II course of the University of Oxford / 2012 / F / F2 / b

The function is smooth, $f(0,0)=0$, andAt the origin both partial derivatives equal one. In particular, $f_y(0,0)=1\ne 0$, so the implicit function theorem supplies a neighbourhood in whichIn fact $h$ is smooth because $f$ is smooth.

By the implicit derivative formula,All sufficiently near points of the curve have $y=h(x)$, and the only such point with $x=0$ is the origin itself. Therefore for the punctured approach along the curve,