Let $F(x,y)=xyI_p$ where $I_p$ is $p$-dimensional identity matrix. I want to calculate
$ \operatorname{tr} \left( F^{-1}\frac{\partial F}{\partial x} F^{-1}\frac{\partial F}{\partial x}\right) \quad $ and $ \quad \operatorname{tr}\left( F^{-1}\frac{\partial F}{\partial x} F^{-1}\frac{\partial F}{\partial y}\right)$
where $\operatorname{tr}( )$ denotes the trace operator. It is straightforward to show that
$\operatorname{tr}\left( F^{-1}\frac{\partial F}{\partial x} F^{-1}\frac{\partial F}{\partial x}\right)=\frac{p}{x^2}, \quad $ and $\quad \operatorname{tr}\left( F^{-1}\frac{\partial F}{\partial x} F^{-1}\frac{\partial F}{\partial y}\right)=\frac{p}{xy}$.
I want to obtain these in another way. Since $d^2\log|F|=-\operatorname{tr}(F^{-1}dFF^{-1}dF)$ and $\log|F|=p\log(x)+p\log(y)$, it follows that
$$\operatorname{tr}\left( F^{-1}\frac{\partial F}{\partial x} F^{-1}\frac{\partial F}{\partial x}\right)=\frac{\partial^2 \log|F|}{\partial x^2}=\frac{p}{x^2}.$$
However, we have
$$\operatorname{tr}\left( F^{-1}\frac{\partial F}{\partial x} F^{-1}\frac{\partial F}{\partial y}\right)=\frac{\partial^2 \log|F|}{\partial x \partial y}=0$$
which is different from the above result. I don't see what is the problem.