= Solution
For a twice continuously differentiable <vector field> in <Cartesian coordinates>, the <curl> and <divergence> are
$$
\nabla\times\mathbf v=(\partial_yv_z-\partial_zv_y,\ \partial_zv_x-\partial_xv_z,\ \partial_xv_y-\partial_yv_x),\qquad\nabla\cdot\mathbf v=\partial_xv_x+\partial_yv_y+\partial_zv_z.
$$
Put $\boldsymbol\omega=\nabla\times\mathbf v$. Using the <Levi-Civita symbol>, the <Kronecker delta> contraction and the <product rule>,
$$
\begin{aligned}
[\nabla\times(\mathbf v\times\boldsymbol\omega)]_i
&=\partial_j(v_i\omega_j-v_j\omega_i)\\
&=\omega_j\partial_jv_i-v_j\partial_j\omega_i-\omega_i\partial_jv_j+v_i\partial_j\omega_j.
\end{aligned}
$$
The last term vanishes because the <divergence of a curl is zero>: the symmetric mixed derivatives contract against an antisymmetric symbol. Thus the <curl of a cross product> identity becomes
$$
\boxed{\nabla\times(\mathbf v\times\boldsymbol\omega)=(\boldsymbol\omega\cdot\nabla)\mathbf v-(\mathbf v\cdot\nabla)\boldsymbol\omega-\boldsymbol\omega(\nabla\cdot\mathbf v).}
$$
For <orthogonal coordinates> $q_i$ with <position vector> $\mathbf r(q_1,q_2,q_3)$, the <scale factors of orthogonal coordinates> and local <unit vectors> are
$$
h_i=\left|\frac{\partial\mathbf r}{\partial q_i}\right|,\qquad\mathbf e_i=\frac1{h_i}\frac{\partial\mathbf r}{\partial q_i},\qquad d\mathbf r=\sum_i h_i\mathbf e_i\,dq_i.
$$
Orthogonality gives $ds^2=\sum_i h_i^2dq_i^2$, and right-handedness fixes $\mathbf e_1\times\mathbf e_2=\mathbf e_3$. In <cylindrical coordinates>, $\mathbf r=(\rho\cos\phi,\rho\sin\phi,z)$, so
$$
\boxed{h_\rho=1,\quad h_\phi=\rho,\quad h_z=1,\qquad \mathbf e_\rho=(\cos\phi,\sin\phi,0),\quad\mathbf e_\phi=(-\sin\phi,\cos\phi,0),\quad\mathbf e_z=(0,0,1).}
$$
The radial vector points away from the axis; the azimuthal vector is tangent in the direction of increasing $\phi$; $\mathbf e_z$ is the Cartesian positive $z$ direction. The azimuthal basis is undefined on the coordinate axis itself.
\Image[/past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2011/paper-1-q1-cylindrical-basis.png]
{title=Cylindrical unit vectors relative to Cartesian axes, viewed from positive z}
For the axial velocity profile, the <cylindrical coordinates> formula gives
$$
\boxed{\boldsymbol\omega=\nabla\times\mathbf u=-\partial_\rho u_z\,\mathbf e_\phi=2u_0\rho\,\mathbf e_\phi.}
$$
The associated <cross product> is
$$
\boxed{\mathbf u\times\boldsymbol\omega=-2u_0^2\rho(1-\rho^2)\mathbf e_\rho.}
$$
This field has only a radial component depending on $\rho$, so its <curl> is zero. Also $\nabla\cdot\mathbf u=\partial_z u_z=0$. The identity above therefore yields
$$
\boxed{(\boldsymbol\omega\cdot\nabla)\mathbf u=(\mathbf u\cdot\nabla)\boldsymbol\omega.}
$$
In fact both <directional derivatives> vanish: the first differentiates $u_z\mathbf e_z$ only in the azimuthal direction, and the second differentiates $\boldsymbol\omega$ only in the $z$ direction. All physical vector fields extend smoothly through $\rho=0$, despite the coordinate-basis singularity.
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