counter example of inverse function theorem Prove that in the inverse function theorem, the hypothesis that $f$ is $C^1$ cannot be weakened to the hypothesis that $f$ is differentiable. I read an example of my teacher, but I can't have any analysis argument for the fact that $f$ is not one to one in any neighborhood of $0$. Here it is,  

$$f(0)=0, \qquad f(x)=x + 2x^2\sin \frac{1}{x}, x\neq 0.$$
Except at $0$, the function is $C^1$, so you can analyse it by examining its derivate on the interval $(0,\varepsilon)$. If $f$ was injective on $(0,\varepsilon)$, then it would have to be monotone and hence its derivative could not have a change of sign there. Show that it does.

Note: I had originally and erroneously looked at the function $f(x)=x + x^2\sin \frac{1}{x}$. That function is indeed bijective, but the set of zeros of $f'$ has an accumulation point at $0$, so it, too, does not yield a diffeomorphism when restricted to any neighbourhood of $0$.
It seems to me that your teacher is wrong. Look here  for a version of the inverse function theorem when the function is only everywhere differentiable but not $C^1$ (see for example this MO question).