Let $M_n(\mathbb R)$ be the set of all real $n \times n$ matrices. Let $S$ be the subset of $M_n(\mathbb R)$ containing all invertible matrices. We consider a map $T:S \to S$ given by
$T(A) = A^{-1}$.
Show that the derivative of $T$ at $A \in S$ is given by $Gradient T(A)(B) = -A^{-1} B A^{-1}$ for $B \in M_n(\mathbb R)$.
Since I couldn't solve the problem, I asked the professor and got a hint:
If you identify $M_n(\mathbb R)$ with $\mathbb R^{n^2}$, then you will see that $S$ can be identified as $\mathbb R^{n^2} \backslash \det^{-1} (0)$, i.e, an open subset of $\mathbb R^{n^2}$. Hence, $Gradient T(A) \in \mathcal L(\mathbb R^{n^2}; \mathbb R^{n^2})$.
I am not sure if your professor's hint is a good one. Usually the derivative of matrix inverse is derived using first-order approximation (see Wikipedia, for instance). In this way the proof is coordinate-free.
Identifying $M_n(\mathbb R)$ with $\mathbb R^{n^2}$, IMHO, only complicates things and it brings along a bunch of indices. If you really want to go this way, see the proof on planetmath.org.