Let $f:\mathbb{R}^3\rightarrow\mathbb{R}^3$ be a smooth vector field. Does there always exist another smooth vector field $g:\mathbb{R}^3\rightarrow\mathbb{R}^3$ such that $\nabla\times g = f$? Naturally, such a field $g$ is not unique as we can add to it any conservative field and find another solution to $\nabla\times g = f$, however this alone does not imply a solution exists for any $f$. This question is analogous to asking in one dimension if all smooth functions have an anti-derivative.
If a $g$ doesn't exist for all $f$, I would like an example of such an $f$ and a proof that no $g$ satisfies the equation $\nabla\times g = f$, however if there always exists such a $g$ for any $f$, I would like a proof of that. Finally, if the result holds and there exists a weaker condition than smoothness, for example, if continuity is sufficient, I would like to know what this condition is.
Note that $$ {\rm div} \circ {\rm rot} = 0 $$ Hence, a field $f$ that can be written as $f = {\rm rot}\, g$ has ${\rm div} \, f = 0$. Therefore $f(x) = (x_1, 0, 0)$ cannot be written as ${\rm rot}\, g$ for any $g$.