= Solution
Embed the planar <vector field> as $\mathbf V=(A(x,y),B(x,y),0)$, independent of $z$. Applying the Cartesian <curl> formula gives
$$
\nabla\times\mathbf V=\left(0,0,\frac{\partial B}{\partial x}-\frac{\partial A}{\partial y}\right).
$$
Consequently, for continuously differentiable components,
$$
\boxed{\nabla\times\mathbf V=0\quad\Longleftrightarrow\quad B_x=A_y}.
$$
This is the local closedness condition for the one-form $A\,dx+B\,dy$. To infer global <path independence> from this condition alone, the domain also matters: it suffices that the planar domain is a <simply connected space>. The polynomial field in the next calculation is defined on the whole plane, so there is no missing-hole obstruction.
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