Be $f(x,y,z)=\mathrm e^{\alpha x+\beta y+\gamma z}$ and $v$ an eigenvector of $\begin{pmatrix} 0 & -\gamma & \beta\\ \gamma & 0 & -\alpha\\ -\beta & \alpha & 0\end{pmatrix}$. Then $f(x,y,z)v$ is an eigenvector of $\rm curl$.
Proof: $\partial_x f(x,y,z) = \alpha f(x,y,z)$, $\partial y f(x,y,z) = \beta f(x,y,z)$, $\partial z f(x,y,z) = \gamma f(x,y,z)$. Therefore the "curl matrix" acts on $f(x,y,z)v$ as if all partial derivatives were replaced by the corresponding factor, which gives the matrix above. Thus, we arrive at an ordinary eigenvector equation, and $v$ (and thus each multiple of $v$) is an eigenvector by assumption.
I'm not sure if those are all eigenvectors, though.
First, writing the vectors in component form (as column vectors) has made it less obvious that you need to use the product rule here.
Let $\hat x, \hat y, \hat z$ be the usual Cartesian basis. The curl takes the form
$$\nabla \times A = (\nabla A^x) \times \hat x + A_x \nabla \times \hat x + \ldots$$
But since $\hat x$ is constant, $\nabla \times \hat x$ is zero. Hence, the formula for curl in Cartesian can be written
$$\nabla \times A = (\nabla A^x) \times \hat x + (\nabla A^y) \times \hat y + (\nabla A^z) \times \hat z$$
Once you do the cross products, you get $(\partial_y A^z - \partial_z A^y) \hat x$ and so on, as you usually would expect.
In spherical, however, the basis vectors depend on position. $\nabla \times \hat \theta$ isn't zero, for instance. That's where the terms in Wikipeida's form come from.
Second, when you converted $3x \hat x - z \hat y + 2 y \hat z$, you converted $x, y, z$ to spherical coordinates, but you didn't convert the basis. It's quite clear that you wrote
$$ 3x \hat x - z \hat y + 2 y \hat z = 3 r \sin \theta \cos \phi \hat x - r \cos \theta \hat y + 2r \sin \theta \sin \phi \hat z$$
You need to write this in terms of the spherical basis vectors in order to apply the formula for curl in spherical properly.
In summary, remember that curl in a general coordinate system is not as simple as it looks in Cartesian. You can always derive the correct formula for a given coordinate basis by using the product rule. If $\alpha$ is a scalar field and $F$ a vector field, then
$$\nabla \times (\alpha F)= (\nabla \alpha) \times F + \alpha \nabla \times F$$
make $\alpha$ the components and $F$ the basis vectors to derive the correct curl formula for your coordinate system.
Second, make sure you write all vectors and vector operators in the same basis.
Best Answer
Eigenfunctions of the curl operator, i.e. $\nabla\times \mathbf B=k\mathbf B$ for eigenvalue $k$, are studied in physics as examples of force-free magnetic fields. Kirk T. MacDonald's notes list many solutions. One of the simplest is Lundquist's solution, which in cylindrical coordinates $(\rho,\phi,z)$ is given by