@ alexsuse , your formula is false. To see that, choose $f(A)=A^{-1}$.
You want the derivative of the function $g:A^{-1}\rightarrow A\rightarrow f(A)$, that is $g=f\circ u$ with $Du_{A^{-1}}:H\rightarrow -AHA$. Thus $Dg_{A^{-1}}:H\in M_n\rightarrow Df_A(-AHA)$. In other words, if $A^{-1}$ is a function of $t$, then $\dfrac{\partial{g}}{\partial{t}}=Df_A(-A\dfrac{\partial{A^{-1}}}{\partial{t}} A)$.