I have got the derivative of a trace w.r.t a real matrix as follows $\frac {\partial}{\partial \mathbf X} tr[\mathbf{(X^TCX)}^{-1}(\mathbf{X^T BX})] = -2 \mathbf{CX(X^TCX)^{-1}X^TBX(X^TCX)^{-1}} + 2\mathbf{BX(X^TCX)^{-1}}$
where $\mathbf {B,C}$ is symmetric.
Now I want to solve the trace w.r.t a complex matrix as follows
$\frac {\partial}{\partial \mathbf X} tr[\mathbf{(X^H CX)}^{-1}(\mathbf{X^H BX})]$
where $\mathbf{B,C}$ conjugate symmertic.
I will feel very gratful if anyone leave me some tips.
Maybe not the final solution but a step forward:
Following calculation is valid only under the assumption that $X$ is square and regular.
You can simplify $(X^H CX)^{-1}X^H BX = X^{-1}C^{-1}X^{-H} X^H B X = X^{-1}C^{-1}B X$.
The Gateaux-derivative $\delta A:=\left.\partial_\varepsilon A(X+\varepsilon \delta X)\right|_{\varepsilon=0}$ of the inverse-matrix operation $A(X):=X^{-1}$ results from
\begin{align*} 1 &= A(X) X,\\ 0 &= \delta(AX) = \delta A\, X + A\,\delta X,\\ \delta A &= -A\,\delta X\,X^{-1}\\ &= \underline{\underline{X^{-1}\,\delta X\,X^{-1}}}. \end{align*}
This helps with the calculation of the Gateaux-derivative of your original equation: \begin{align*} \delta \operatorname{tr}[X^{-1}C^{-1}B X] &= \operatorname{tr}[X^{-1} \delta X X^{-1} C^{-1}B X + X^{-1}C^{-1}B \delta X] \end{align*}