= Solution
The derivative <matrix>
$$
\partial_jF_k=\omega_k\omega_j-|\boldsymbol\omega|^2\delta_{kj}
$$
is symmetric in $j,k$. Contracting it with the antisymmetric <Levi-Civita symbol> gives the <curl>
$$
(\nabla\times\mathbf F)_i=\epsilon_{ijk}(\omega_k\omega_j-|\boldsymbol\omega|^2\delta_{kj})=0.
$$
Direct differentiation of the <scalar potential> gives
$$
\partial_i\phi=\omega_i(\boldsymbol\omega\cdot\mathbf x)-|\boldsymbol\omega|^2x_i=F_i,
$$
so \b[$\mathbf F=\nabla\phi$]. To identify the <level sets>, use the <cross product> identity
$$
\phi=-\tfrac12|\boldsymbol\omega\times\mathbf x|^2.
$$
If $\boldsymbol\omega\ne0$, write $\mathbf x=\mathbf x_{\parallel}+\mathbf x_{\perp}$ relative to its direction. Then $\phi=-|\boldsymbol\omega|^2|\mathbf x_{\perp}|^2/2$. Thus \b[each negative level $\phi=c$ is a <circular cylinder> with axis $\mathbb R\boldsymbol\omega$ and radius $\sqrt{-2c}/|\boldsymbol\omega|$]. The zero level is the axis itself; positive levels are empty. If $\boldsymbol\omega=0$, the <scalar potential> is identically zero, so its zero level is all of space and every other level is empty.
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