No assumption on the square matrix $H=(h_{ij})$ are needed. If $\Sigma:=\exp(-\rho H)$, then $\Sigma^{-1}=\exp(\rho H)$ with derivative wrto $\rho$ equal to $(\Sigma^{-1})'=H \exp(\rho H)$. By the Liouville formula for the solution of linear system, $\big(\det(\Sigma)\big)'=-(\operatorname{tr}H)\det(\Sigma) $, and $\big(\log \det(\Sigma)\big)'=-(\operatorname{tr}H).$

[modified according to the clarification given in comments].

In general, for an invertible square matrix $\Sigma=\Sigma(\rho)$, differentiably depending on the real variable $\rho$, we have: $(\Sigma^{-1})'=-\Sigma^{-1} \Sigma' \Sigma^{-1}$, and $\big(\det(\Sigma)\big)'=\operatorname{tr} (\Sigma^{-1} \Sigma')\det(\Sigma)$, so that $\big(\log \det(\Sigma)\big)'=\operatorname{tr} (\Sigma^{-1} \Sigma')$.

Here, $\Sigma'$ is the matrix with entries $-h_{ij}e^{-\rho h_{ij}}$, the Hadamard product of the matrices $-(h_{ij})_{ij}$ and $\Sigma$; and since both $\Sigma^{-1}$ and $\Sigma'$ are symmetric, $\operatorname{tr} (\Sigma^{-1} \Sigma')$ is just their Frobenius scalar product. I do not see other simplifications, unfortunately.