= Solution
In <Cartesian coordinates>, independent coordinate variables satisfy
$$
\frac{\partial x_i}{\partial x_j}=\delta_{ij},\qquad
\frac{\partial r}{\partial x_j}=\frac{x_j}{r}\quad(r>0),
$$
where $\delta_{ij}$ is the <Kronecker delta>. Applying the <product rule> and <chain rule> to $u_i=r^\alpha x_i$ gives the <Jacobian matrix>
$$
\boxed{d_{ij}=r^\alpha\delta_{ij}+\alpha r^{\alpha-2}x_ix_j.}
$$
This <tensor> is symmetric in $i,j$. Hence its contraction with the antisymmetric <Levi-Civita symbol> vanishes:
$$
(\nabla\times\mathbf u)_i=\epsilon_{ijk}d_{kj}=0.
$$
Similarly, $v_i=\epsilon_{ijk}k_ju_k$ and the constancy of $\mathbf k$ give
$$
\nabla\cdot\mathbf v=\epsilon_{ijk}k_jd_{ki}=0.
$$
The trace of the <Jacobian matrix> is
$$
\nabla\cdot\mathbf u=(3+\alpha)r^\alpha,
$$
which vanishes for $\alpha=-3$. The <curl of a cross product> identity, with $\mathbf k$ constant, now gives
$$
\nabla\times(\mathbf k\times\mathbf u)
=\mathbf k(\nabla\cdot\mathbf u)-(\mathbf k\cdot\nabla)\mathbf u.
$$
For $\alpha=-3$,
$$
(\mathbf k\cdot\nabla)\mathbf u=r^{-3}\mathbf k-3r^{-5}(\mathbf k\cdot\mathbf x)\mathbf x,
$$
so
$$
\boxed{\nabla\times\mathbf v=\frac{3(\mathbf k\cdot\mathbf x)\mathbf x-r^2\mathbf k}{r^5}.}
$$
All identities for this singular case hold on $r>0$. In particular, zero <divergence> here is a pointwise assertion away from the origin, not an assertion that the flux through a sphere enclosing the singularity is zero.
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