I know that $\nabla_\sigma g_{\mu\nu}=0$, being $\nabla_\sigma$ the covariant derivative, due to metric compatibility. I've also seen somewhere else that $\nabla_\sigma g=0$ as well, being $g$ the determinant of the metric tensor $g_{\mu\nu}$. Nonetheless, I don't know if the property:
$\nabla\left(S\otimes T\right)=\left(\nabla S\right)\otimes T+S\otimes\left(\nabla T\right)$
Holds in the case of the determinant, and therefore if $\nabla_\sigma\sqrt{g}g_{\mu\nu}=0$ for all values of $g_{\mu\nu}$.
Since you know that metric compatibility of the connection implies $$ \nabla_\sigma\,g_{\mu\nu}=0 $$ we only have to examine $\nabla_\sigma\sqrt{g}$ which is a partial derivative since $\sqrt{g}$ is a scalar. By the product rule, \begin{align}\require{cancel} \nabla_\sigma(\sqrt{g}\,g_{\mu\nu})=(\partial_\sigma\sqrt{g})g_{\mu\nu}+\cancel{\sqrt{g}\,\nabla_\sigma\,g_{\mu\nu}}\,. \end{align} This is zero, if either the component $g_{\mu\nu}$ is zero, or if $\partial_\sigma\sqrt{g}$ is zero. To figure out when the later holds we use one of the very useful formulas for the contracted Christoffel symbol $$ {\Gamma^\color{red}{\rho}}_{\color{red}{\rho}\,\sigma}=\partial_\sigma\log\sqrt{g}=\frac{\partial_\sigma\sqrt{q}}{\sqrt{q}}\,. $$