I am trying to find the $\textit{second}$ derivative of $X \mapsto (F \circ G) (X)$, where $G(X)=\frac{AXA}{\operatorname{tr}(AXA)}$ and $F(X)=\operatorname{tr}\left((CXC)^\frac{1}{2}\right)$ are functions of the positive matrix X. The matrices A and C are constant and also positive.
I computed the first derivative and got
$\frac{\partial (F \circ G)}{\partial X}=\frac{1}{2}\operatorname{tr}\left( AXA\right)^{-1}\left(AC(CG(X)C)^{-\frac{1}{2}}CA -\operatorname{tr} \left((CG(X)C)^\frac{1}{2} \right)A^2\right)$
notably thanks to:
Derivative of $\mbox{Tr} \left(\left(AXA\right)^{\frac{1}{2}}\right)$ with respect to $X$
Derivative of $X \mapsto \frac{AXA}{\operatorname{Tr} \left( AXA \right )}$
However, to derive again, I am stuck with the inverse square root $(CG(X)C)^{-\frac{1}{2}}$. Would you know how to deal with it?