= Solution
In <Cartesian coordinates>, use the <Levi-Civita symbol> and the <Einstein summation convention> to expand
$$
\begin{aligned}
[\mathbf u\times(\nabla\times\mathbf u)]_i
&=\epsilon_{ijk}\epsilon_{klm}u_j\partial_lu_m\\
&=u_j\partial_i u_j-u_j\partial_j u_i\\
&=\frac12\partial_i(u_ju_j)-[(\mathbf u\cdot\nabla)\mathbf u]_i.
\end{aligned}
$$
The contraction of the two symbols uses the <Kronecker delta>, and the final step is the <product rule>. Thus the <convective acceleration> identity is
$$
\boxed{(\mathbf u\cdot\nabla)\mathbf u=\frac12\nabla|\mathbf u|^2-\mathbf u\times(\nabla\times\mathbf u).}
$$
Put $\boldsymbol\omega=\nabla\times\mathbf u$. The <curl of a gradient> vanishes, and the <curl of a cross product> identity gives
$$
\nabla\times(\mathbf u\times\boldsymbol\omega)=\mathbf u\,\nabla\cdot\boldsymbol\omega-\boldsymbol\omega\,\nabla\cdot\mathbf u+(\boldsymbol\omega\cdot\nabla)\mathbf u-(\mathbf u\cdot\nabla)\boldsymbol\omega.
$$
Since the <divergence of a curl is zero>, taking the <curl> of the first identity yields
$$
\boxed{\nabla\times[(\mathbf u\cdot\nabla)\mathbf u]=(\nabla\cdot\mathbf u)\boldsymbol\omega+(\mathbf u\cdot\nabla)\boldsymbol\omega-(\boldsymbol\omega\cdot\nabla)\mathbf u.}
$$
Twice continuous differentiability suffices for the mixed-derivative identities used here.
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