Consider the mapping $f: \mathbb R^ 3 \to \mathbb R ^3$ defined by $f(x,y,z) = (x, y^3, z^5)$ Consider the mapping $f: \mathbb R^ 3 \to \mathbb R ^3$ defined by $f(x,y,z) = (x, y^3, z^5)$.
$f$ has a (global) inverse $g$, despite that the matrix $f '(0,0,0)$ is singular. What does this imply about the differentiability of $g$ at $(0,0,0)$.  
Thanks for any assistance.

For an easier example, just think about $f: \Bbb R \mapsto \Bbb R$ given by $f(x)=x^{3}$, it also has an inverse, even though $f'(0)$ is zero. What is the inverse and is it differentiable at $0$?
