Consider the anisotropic heat equation in $\mathbb{R}^n$: $$ \partial_tu=\partial_i(g^{ij}\partial_j u) $$ where $\displaystyle \partial_i = \partial_{x_i}$ and $g^{ij}$ are the components of the inverse of some metric tensor.
I want to ask when we also have: $$ \partial_t u = \Delta_g u $$ whenever the first equation holds.
I can do this if $\det(g)=|g|$ is constant: \begin{align} \partial_tu &= \partial_i(g^{ij}\partial_j u)\\ &= \partial_ig^{ij}\partial_j u + g^{ij}\partial_{ij}u\\[1mm] &= \sqrt{|g|}^{-1}\partial_i\left(\sqrt{|g|}g^{ij}\right)\partial_j u + g^{ij}\partial_{ij}u \\ &= \sqrt{|g|}^{-1}\left[ -\sqrt{|g|}g^{i\ell}\Gamma^j_{i\ell} \right] \partial_j u + g^{ij}\partial_{ij}u \\ &= \Delta_g u \end{align}
where I used the identities: $$ g^{i\ell}\Gamma^j_{i\ell}=\frac{-1}{\sqrt{|g|}}\partial_i\left(g^{ji}\sqrt{|g|}\right) $$ $$ \Delta_g = g^{jk}\partial_{jk} - g^{jk}\Gamma_{jk}^\ell\partial_\ell $$ Or, as I just realized, more easily, using the identity: $\Delta_g=\sqrt{|g|}^{-1}\partial_i(\sqrt{|g|}g^{ij}\partial_j) $.
Does this hold in other cases?
As mentioned in the comments, you can see that: $$ \Delta_g u= g^{ij}\partial_{ij} u + \left( g^{ij}\partial_i\sqrt{|g|}+\partial_i g^{ij} \right)\partial_j u $$ So we have $$ \Delta_g u = g^{ij}\partial_{ij}u+\partial_i g^{ij}\partial_j u $$ only if $$ g^{ij}\partial_i\sqrt{|g|}\partial_j u =0 $$ Clearly, this is true if $|g|$ is constant.
Just to expand this a bit, we see that it is equivalent to $$ \frac{1}{2}g^{ij}\sqrt{|g|}\;\text{tr}(g^{-1}\partial_i g)\partial_j u = 0 $$ using the formula for the derivative of a determinant.
Anyone who has an interpretation for this is welcome to comment on it.