Let $(M, g)$ be a Riemannian manifold, $f: M \rightarrow \mathbb{R}$, and $h$ a vector field on $M$ that vanishes on the boundary and which is divergence free. I would like to show that, using coordinates, $$\int (\partial_i f) h^j g^{ij} dV = 0$$ where $dV = \sqrt{\det(g_{ij})} dx^1 \wedge \cdots \wedge dx^n$ is the Riemannian volume form. Thus the integral becomes $$\int (\partial_i f) h^j g^{ij} \sqrt{\det(g_{ij})} dx = 0 \tag{1}$$ where I have written $dx = dx^1\cdots dx^n$ for convenience.
In local coordinates the divergence takes the form $$Div(h) = \frac{1}{\sqrt{\det(g_{ij})}}\partial_i\Big(h^i\sqrt{\det(g_{ij})}\Big).$$
My first approach was to integrate (1) by parts to get $$\int (\partial_i f) h^j g^{ij} \sqrt{\det(g_{ij})} dx = -\int f(\partial_ih^j)g^{ij}\sqrt{\det(g_{ij})} dx -\int fh^j\partial_i(g^{ij})\sqrt{\det(g_{ij})} dx -\int fh^jg^{ij}\partial_i\big(\sqrt{\det(g_{ij})}\big) dx \\ = -\int fh^j\partial_i(g^{ij})\sqrt{\det(g_{ij})} dx - \int fg^{ij}\partial_i\Big(h^j\sqrt{\det(g_{ij})}\Big) dx $$ The last term looks a lot like the definition of the divergence other than the factor of the square root but the index on the partial derivative does not match the index of $h$. I also do not see how the first term after the last inequality can be made to vanish without anymore information.
Does anyone know how I can proceed?