= Solution
Use <suffix notation>, summing repeated indices. The <contraction of two Levi-Civita symbols> gives
$$
[F\times(\nabla\times G)]_i=\epsilon_{ijk}F_j\epsilon_{k\ell m}\partial_\ell G_m=F_j\partial_iG_j-F_j\partial_jG_i.
$$
Adding the <directional derivative> $(F\cdot\nabla)G_i=F_j\partial_jG_i$ cancels the second term. Interchanging $F,G$ gives another pair whose sum is $G_j\partial_iF_j$. The total right-hand side is therefore
$$
F_j\partial_iG_j+G_j\partial_iF_j=\partial_i(F_jG_j),
$$
the $i$th component of the <gradient> of the <dot product>. Since this holds for each component, the <dot-product gradient identity> is proved:
$$
\boxed{\nabla(F\cdot G)=(F\cdot\nabla)G+(G\cdot\nabla)F+F\times(\nabla\times G)+G\times(\nabla\times F)}.
$$
It holds for any continuously differentiable <vector fields>; no irrotationality assumption is used.
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