Using the divergence theorem, we can compute volume by integrating along the surface: $$\mathrm{vol}(M)=\int_M\mathrm{d}V=\oint_{\partial M}\mathrm{d}S \,\vec{n}\cdot\vec{v},$$ where $\vec{n}$ is normal to $\partial M$ and $\vec{v}$ is a vector fulfilling $\vec{\nabla}\cdot\vec{v}=1$, e.g. $\vec{v}=x\hat{e_x}$. Bringing it into the from $$\mathrm{vol}(M)=\oint_{\partial M}\mathrm{d}^{d-1}x\,n_\alpha x^\alpha\sqrt{g},$$ where now $g$ is the determinant of the metric tensor and $n_\alpha$ the normal to $\partial M$, I wanted to compute the variation of this integral w.r.t. the metric $g_{ij}$: $$\oint_{\partial M}\mathrm{d}^{d-1}x\,\left(\frac{1}{\sqrt{g}}\frac{\delta(n_\alpha x^\alpha\sqrt{g})}{\delta g^{ij}}\right)\sqrt{g}.$$ Using the product rule, this boils down in computing the first term in $$\frac{1}{\sqrt{g}}\frac{\delta(n_\alpha x^\alpha\sqrt{g})}{\delta g^{ij}}=\frac{1}{\sqrt{g}}\frac{\delta(n_\alpha x^\alpha)}{\delta g^{ij}}-\frac{n_\alpha x^\alpha}{2}g_{ij},$$ but I'm not sure how to do it. $x^\alpha$ are components of the parametrization of $\partial M$.
Edit: In this question the functional derivative of $n_\alpha$ is given as (after lowering/raising indices) $$\frac{\delta n_\alpha}{\delta g^{ij}}=-\frac{1}{2}n_\alpha n_in_j$$ but no derivation is stated nor where to look it up. The only other quantity missing then is $$\frac{\delta x^\alpha}{\delta g^{ij}}=?$$
Edit 2: As a follow up question, what is the functional derivative of the mean curvature $$H=-\frac{1}{2}\nabla_\alpha n^\alpha$$ w.r.t. the metric, i.e. $$\frac{\delta H}{\delta g^{ij}}=?.$$ This is related to the derivative of the normal, as we have $$H=-\frac{1}{2}\nabla_\alpha n^\alpha=-\frac{1}{2\sqrt{g}}\frac{\partial(\sqrt{g}n^\alpha)}{\partial x^\alpha}$$ and thus $$\frac{\delta H}{\delta g^{ij}}=\frac{H}{4}g_{ij}-\frac{1}{2\sqrt{g}}\frac{\delta}{\delta g^{ij}}\left(\frac{\partial(\sqrt{g}n^\alpha)}{\partial x^\alpha}\right),$$ but I'm not sure what to do about the last term.
