An additional assumption to the inverse function theorem. The theorem is given below:




And here is the question:

Could anyone give me a hint on how to prove the required in the question please?
 A: There is a more general form of Cramer's rule than the one used for solving system's of linear equations. See the section on inverting matrices here on Wikipedia for example: https://en.wikipedia.org/wiki/Cramer%27s_rule#Finding_inverse_matrix. 
In this question in particular we have that $Dh(v)= (Df(h(v)))^{-1}$ by part (e) of the inverse function theorem. To show that the inverse function is also a $\mathcal{C}^{r}$ function we must show that the partial derivatives $\frac{\partial h_i}{\partial x_j}$ of the function are $\mathcal{C}^{r-1}$ (as then the original function will be $\mathcal{C}^r$.)
Using the version of Cramer's rule above we have that the ij-th entry of $(Df(h(v)))^{-1}$ (and hence the partial derivative $\frac{\partial h_i}{\partial x_j}$) will be given by: 
$$
((Df(h(v))^{-1})_{ij} = \frac{\det(A)}{\det(Df(v(h)))}
$$
Where $A$ is the matrix obtained from $Df(v(h))$ by co-factor expanding on the ij-th entry of the matrix.
This expression gives $\frac{\partial h_i}{\partial x_j}$ in terms of the partial derivatives of $f$. In particular we see that the partial derivative of $h_i$ with respect to $x_j$ is a quotient of a sum and product of $\mathcal{C}^{r-1}$ functions and a no-where zero $\mathcal{C}^{r-1}$ function on the ball. 
We conclude from this that the partial derivates of the inverse function are $\mathcal{C}^{r-1}$ and hence the inverse function is $\mathcal{C}^r$.
