On the invertible locus, $D\mathrm{inv}(A)[H]=-A^{-1}H{A}^{-1}$

Because $A$ is invertible, $\det (A)\mathrm{\ne }0$. Since $\det (A+K)$ is a polynomial in the real coordinates of $K$ with value $\det (A)$ at $K=0$, there is $\epsilon >0$ such that $A+K$ is invertible whenever $\mathrm{\parallel }K{\mathrm{\parallel }}_F<\epsilon$. For such $K$, the identities $(A+K{)}^{-1}(A+K)=I=A^{-1}A$ give $(A+K{)}^{-1}-A^{-1}=-A^{-1}K(A+K{)}^{-1},$ and hence $((A+K{)}^{-1}-A^{-1})+A^{-1}K{A}^{-1}=A^{-1}K{A}^{-1}K(A+K{)}^{-1}\mathrm{.}$

Shrink $\epsilon$ so that $\mathrm{\parallel }{A}^{-1}{\mathrm{\parallel }}_2\mathrm{\parallel }K{\mathrm{\parallel }}_F\le \frac{1}{2}$ whenever $\mathrm{\parallel }K{\mathrm{\parallel }}_F<\epsilon$. From step 1.1, $\mathrm{\parallel }(A+K{)}^{-1}{\mathrm{\parallel }}_2\le \mathrm{\parallel }{A}^{-1}{\mathrm{\parallel }}_2+\mathrm{\parallel }{A}^{-1}{\mathrm{\parallel }}_2\mathrm{\parallel }K{\mathrm{\parallel }}_F\mathrm{\parallel }(A+K{)}^{-1}{\mathrm{\parallel }}_2,$ so $\mathrm{\parallel }(A+K{)}^{-1}{\mathrm{\parallel }}_2\le 2\mathrm{\parallel }{A}^{-1}{\mathrm{\parallel }}_2$. Applying this bound to the second identity in step 1.1 yields $\parallel (A+K{)}^{-1}-A^{-1}+A^{-1}K{A}^{-1}{\parallel }_F\le 2\mathrm{\parallel }{A}^{-1}{\mathrm{\parallel }}_2^3\mathrm{\parallel }K{\mathrm{\parallel }}_F^2=o(\mathrm{\parallel }K{\mathrm{\parallel }}_F)\mathrm{.}$ Therefore inversion is real Fr'echet differentiable at $A$ with derivative $K\mapsto -A^{-1}K{A}^{-1}$.

Differentiating the identity $A{A}^{-1}=I$ and using the product rule [L1] gives $H{A}^{-1}+AD\mathrm{inv}(A)[H]=0.$ Left-multiplying by $A^{-1}$ recovers $D\mathrm{inv}(A)[H]=-A^{-1}H{A}^{-1},$ which is the formula claimed in the statement.