Let $A$ be a symmetric positive definite matrix. Let the determinant of $A$ be given by $|A|$.
How to find the following gradient \begin{align} \nabla_{vec(A)} |A| \end{align} that is how to find gradient of $|A|$ with respec to $vec(A)$ where $vec(\cdot)$ is the vectorization operator.
$\def\p#1#2{\frac{\partial #1}{\partial #2}}\def\v{{\rm vec}}$Use the well-known Jacobi formula to calculate the differential with respect to the matrix itself $$\eqalign{ d\det(A) &= \det(A)\,A^{-T}:dA \\ &= \det(A)\,A^{-1}:dA \qquad \big(A\,\,{\rm is\,\,symmetric}\big) \\ }$$ Vectorize this expression to calculate the differential with respect to the vector, then construct the desired gradient. $$\eqalign{ d\det(A) &= \det(A)\,\v(A^{-1}):\v(dA) \\ \p{\det(A)}{\v(A)} &= \det(A)\,\v(A^{-1}) \\\\ }$$
In the above, a colon is used as a convenient product notation for the trace, i.e. $$\eqalign{ A:B = {\rm Tr}(A^TB) \;=\; \sum_{i=1}^m \sum_{j=1}^n A_{ij} B_{ij} \\ }$$