Given a scalar function $f:{\mathbb R}\to{\mathbb R},\;$ its derivative
$f'=\frac{df}{dx}$ and the following definitions
$$\eqalign{
v &= f(\mu),\quad &v' = f'(\mu) \qquad &\big({\rm functions\,applied\,elementwise}\big) \\
V &= {\rm Diag}(v),\quad &V' = {\rm Diag}(v') \\
\phi &= \log\big|\!\det(V-&VAV)\big| \\
}$$
Then the gradient of the $\phi$-function wrt $\mu$ is
$$\eqalign{
\frac{\partial\phi}{\partial\mu}
&= V'\operatorname{diag}\Big(X^{-1}-AVX^{-1}-X^{-1}VA\Big) \\
}$$
where the diag() function returns the main diagonal of its matrix argument as a column vector, while the Diag() function creates a diagonal matrix from a vector argument.
The derivation of this gradient follows.
The differential of the element-wise vector function is
$\; dv = V'\,d\mu$
Define the matrix $\;X = (V-VAV),\;$
write the $\phi$-function in terms of it,
utilize the result from the Matrix Cookbook,
and then perform a change of variables from $X\to V\to v\to\mu$.
$$\eqalign{
\phi &= \log(\det(X)) \\
d\phi &= X^{-T}:dX \\
&= X^{-T}:\big(dV-dV\,AV-VA\,dV\big) \\
&= \big(X^{-T}-X^{-T}(AV)^T-(VA)^TX^{-T}\big):dV \\
&= \big(X^{-1}-AVX^{-1}-X^{-1}VA\big):dV \\
&= \big(X^{-1}-AVX^{-1}-X^{-1}VA\big):{\rm Diag}(dv) \\
&= {\rm diag}\big(X^{-1}-AVX^{-1}-X^{-1}VA\big):dv \\
&= {\rm diag}\big(X^{-1}-AVX^{-1}-X^{-1}VA\big):V'\,d\mu \\
&= V'\operatorname{diag}\Big(X^{-1}-AVX^{-1}-X^{-1}VA\Big):d\mu
}$$
which recovers the gradient shown above.
In the steps above, a colon denotes the trace/Frobenius product, i.e.
$$A:B = {\rm Tr}(A^TB)$$
The cyclic property of the trace allows the terms in such a product
to be rearranged in many different ways, e.g.
$$\eqalign{
A:B &= A^T:B^T &= B:A \\
A:BC &= B^TA:C &= AC^T:B \;= I:A^TBC \;= etc \\
}$$
Several steps also made use of the fact that
$(V,V')$ are symmetric matrices.
