Consider the equation $$\frac{\partial \vec\omega}{\partial t}+(\vec u \cdot \nabla)\vec \omega=(\vec\omega \cdot \nabla)\vec u $$ where $\vec \omega = \nabla \times \vec u$. If $\vec u$ is 2-dimensional, i.e., $\vec u = (u_1, u_2, 0)$ and $\partial \vec{u} / \partial x_3 = 0$, then show that $\vec \omega $ is constant.
Clearly I need to show that $\partial \vec \omega/\partial t=0$ but I'm unsure of how to show that $(\vec u \cdot \nabla)\vec \omega$ and $(\vec\omega \cdot \nabla)\vec u$ are zero.
You're misinterpreting "constant" - it must mean that the material derivative $d\omega / dt = 0$, not that $\partial \omega/\partial t = 0$. Thus the vorticity will be constant if measured by an observer moving with the fluid, but not if measured at a fixed point. (For a counterexample to $\partial_t \omega = 0$, consider the field $u=(x y, - y^2 / 2,0)$, which even satisfies $\nabla \cdot u = 0$.)
Since $$\frac{d \omega }{ d t }= \frac{\partial \omega }{ \partial t }+ (u \cdot \nabla) \omega,$$ to show that $d \omega / dt = 0$ from the given evolution equation we just need to show that $(\omega \cdot \nabla) u = 0$. This is simple: since $u$ is two-dimensional, the only non-zero partial derivatives $\partial_i u_j$ are when both $i,j$ are $1$ or $2$, so the curl must be in the $x_3$ direction; but we know by assumption that $u$ is constant in the $x_3$ direction.