 Solution

Differentiability at $a$ means that a linear map $L:V\to V_1$ satisfiesThis map is unique and is the total derivative $D{f}_a$. In finite dimensions linear maps are continuous.

The Frobenius norm is submultiplicative. Expanding the noncommuting cube givesThe last four terms have norm at most $3|A||H{|}^2+|H{|}^3=o(|H|)$. ThusIf $|C|<1$, the Neumann series ${\sum }_{j\ge 0}(-C{)}^j$ converges and is the inverse of $I+C$, as follows by multiplying its partial sums and letting the final power tend to zero. Its terms from degree two onwards have norm at most $|C{|}^2/(1-|C|)$. Hence $(I+C{)}^{-1}=I-C+O(|C{|}^2)$.

For invertible $A$, write $A+H=A(I+A^{-1}H)$. Whenever $|A^{-1}||H|<1$, this is invertible, proving the general linear group is open. The same expansion givesCompose this inverse derivative with the cube derivative, setting $B=A^{-1}$. The result isThe three terms cannot generally be combined into a scalar-style $-3A^{-4}H$, because matrix factors need not commute.

This is the cubic case of the Fréchet derivative of a negative matrix power.