For some vector field, $<P, Q>$, is it possible for $P_{y} = Q_{x}$ and yet not have $\nabla f = <P,Q>$

Question

Even if $P_{y} = Q_{x}$, is it guaranteed that some function exists whose gradient gives $<P,Q>$?

Answer 1

The vector field

$$<P,Q>= \langle -\frac{y}{x^2+y^2},\frac{x}{x^2+y^2} \rangle $$

satisfies $P_y=Q_x$. However, it is not the gradient of a function defined in $\mathbb{R}^2 \setminus \{(0,0) \}.$

Answer 2

A closed differential form (or vector field) must be defined and bounded in a star-shaped region for it to be exact. The example user1337 gives fails because it’s undefined at the origin, so the region where it is defined—$\mathbb{R}^2\setminus\{(0,0)\}$—isn’t star-shaped.