Let $\phi$,$\lambda$ be scalar fields
R.T.P $$\operatorname{div}(\vec{\nabla} \phi \times \vec{\nabla} \lambda) = 0$$ using index notation
what I had done was the following:
\begin{align*} \operatorname{div}(\vec{\nabla} \phi \times \vec{\nabla} \lambda) &= \operatorname{div} (\epsilon_{ijk} \cdot \vec{e_{i}} \cdot (\nabla \phi)_{j} \cdot (\nabla \lambda)_{k})\\ &= \operatorname{div}(\epsilon_{ijk} \cdot \vec{e_{i}} \cdot \phi_{,j} \cdot \lambda_{,k})\\ &= \operatorname{div}(\epsilon_{ijk} \cdot \phi_{,j} \cdot \lambda_{,k})\\ &= (\epsilon_{ijk} \cdot \phi_{,j} \cdot \lambda_{,k})_{,i}\\ &= \epsilon_{ijk,i} \cdot (\phi_{,j} \cdot \lambda_{,k}) + \epsilon_{ijk} \cdot (\phi_{,j} \cdot \lambda_{,k})_{,i}\\ &=\epsilon_{ijk} \cdot (\phi_{,j} \cdot \lambda_{,k})_{,i}\\ &= \epsilon_{ijk} (\phi_{,ji} \cdot \lambda_{,k} + \phi_{,j} \cdot \lambda_{,ki} ) \end{align*}
from here I do not know how to proceed any further. is there an identity I could use?