= Solution
Differentiability at $a$ means that a linear map $L:V\to V_1$ satisfies
$$
\frac{\|f(a+h)-f(a)-Lh\|_1}{\|h\|}\longrightarrow0\quad(h\to0).
$$
This map is unique and is the total derivative $Df_a$. In finite dimensions linear maps are continuous.
The <Frobenius norm> is submultiplicative. Expanding the noncommuting cube gives
$$
(A+H)^3=A^3+A^2H+AHA+HA^2+AH^2+HAH+H^2A+H^3.
$$
The last four terms have <norm> at most $3|A||H|^2+|H|^3=o(|H|)$. Thus
$$
\boxed{Df_A(H)=A^2H+AHA+HA^2}.
$$
If $|C|<1$, the <Neumann series> $\sum_{j\ge0}(-C)^j$ 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 gives
$$
(A+H)^{-1}=A^{-1}-A^{-1}HA^{-1}+O(|H|^2).
$$
Compose this inverse derivative with the cube derivative, setting $B=A^{-1}$. The result is
$$
\boxed{Dg_A(H)=-(A^{-3}HA^{-1}+A^{-2}HA^{-2}+A^{-1}HA^{-3})}.
$$
The 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>.
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