Differential of an inverse function - do we need continuity? The theorem about the derivative of the inverse function says that if 


*

*$f$ is differentiable with $f(x_0) \neq 0$

*$f$ has an inverse function

*$f^{-1}$ is continuous in $f(x_0)$
then $f^{-1}$ is differentiable in $y = f(x_0)$


Do we need the assumption about the continuity of the inverse function?
 A: You forgot to mention that $f'(x_0)$ must be non singular. With this assumption, and the mild extra assumption of continuity of $f'$ in a neighborhood of $x_0$, the inverse function theorem kicks in and automatically ensures continuity and differentiability of the inverse function. (This is true even in the general context of Banach spaces).

This leaves open the existence of a differentiable functions $f$ with non-continuous derivative $f'$, with the property that $f'(x_0)$ is nonsingular and the inverse function $f^{-1}$ is not differentiable at $f(x_0)$. 

I am not sure you can find such examples in finite dimension. In infinite dimension, however, those abound: consider for instance the injection $I\colon \ell^1\to \ell^2$, which is linear and continuous, hence differentiable with nonsingular differential. The inverse function $I^{-1}\colon I(\ell^1) \to \ell^1$ is not continuous.
(I had written some personal notes about this topic here. Those are meant for my personal record, but perhaps they can be of some help. Lemma 1 is the reason why I suspect one cannot find examples of functions with discontinuous inverse in finite dimension.)
