One example: $f'(a)$ isn't invertible function, then $(f^{-1})'(f(a))$ isn't invertible. I would like to find one example: let be $f:A\subset \mathbb{R}^p \to \mathbb{R}^p$ and  $g:f(A) \to \mathbb{R}^p$ its inverse function. Supose that $f$ is differentiable in $a\in A$ and $g$ in $b= f(a)$. I would like to show that if $f'(a)$ isn't invertible function, then $g'(b)$ isn't invertible.
I tried $f(x)=x^3$ and $a = 0$, but the inverse function is not diferentiable in $0=f(0)$. 
 A: The general conditions for the existence of the inverse of any differentiable function f: $\mathbb R^n \rightarrow \mathbb R^m$ are given by the inverse function theorem. I'll state the theorem without proof-see any rigorous multivariable calculus text such as Bandaxall and Liebeck or Munkres for the proof: 
Inverse Function Theorem in $\mathbb R^n$ Let f : $\mathbb R^n \rightarrow \mathbb R^m$ such that 
1) f is continuously differentiable from an open set U $\subseteq \mathbb R^n$ to an open set V$\subset\mathbb R^m$ .
2) The total derivative $D_f$ of f is defined at a point p $\in U$.
3) $D_f$(p) is invertible i.e. the Jacobian of f(p) is invertible in an open neighborhood of  p.
Then if all 3 of these conditions are met, then there exists a neighborhood of  p where f(p) is invertible and the inverse of f exists in a neighborhood of f(p) and $f^{-1}$ is also continously differentiable at f(p). 
Note what he theorem does not say. It does not say if a function is locally invertible,then it's invertible on it's entire domain. Note also the example you're trying to set up is actually the negation of (3).In other words,you want a counterexample where (1) and (2) hold,but (3) doesn't.The IFT seems to imply that this counterexample does not exist.
If the derivative (either ordinary where n=1 or the total derivative when n > 1) isn't invertable, then the theorem implies f doesn't have an inverse either. 
If someone else produces a counterexample, I'd be really surprised. 
A: If $f$ is differentiable at $a$ and $g$ at $b=f(a)$, then applying the chain rule to $Id_A = g\circ f$ we get that
$$ D_g(b)D_f(a)=I_p,$$
where $I_p$ is the $p\times p$ identity matrix. This means that both $D_g(b)$ and $D_f(a)$ have to be invertible if we assume that both of them exist.
