My approach was to assign $dv =\frac{\partial{v}}{\partial{x}}dx + \frac{\partial{v}}{\partial{y}}dy + \frac{\partial{v}}{\partial{z}}dz$ , $du =\frac{\partial{u}}{\partial{x}}dx + \frac{\partial{u}}{\partial{y}}dy + \frac{\partial{u}}{\partial{z}}dz$ , $u=u$, and $v=v$
then using the partial fraction formula: $$\oint{v \nabla u d\vec{r}} = u.v - \int{v du} = u.v - \int{v . \nabla u d\vec{r}} = -v\nabla ud\vec{r}$$
this assumes that $ du = \nabla u .d \vec{r} $ and I am not sure if that assumption is corrrect as I am unsure if me assuming $\iiint \frac{\partial{u}}{\partial{x}}dx + \frac{\partial{u}}{\partial{y}}dy + \frac{\partial{u}}{\partial{z}}dz= u$ is correct.
Can I get any insight if what I am doing is incorrect and maybe a hint on how to apprroach this correctly? Thank you.
It follows from the fundamental theorem of line integrals that $\oint \nabla(uv) d\vec{r} = 0$. Since $\nabla(uv) = u \nabla(v) + v \nabla(u)$, you get the result by integrating both sides.