Let $\Omega \subseteq \Bbb R^2$ be an open set, and let $\omega = \omega_1\,{\rm d}x_1 +\omega_2 \,{\rm d}x_2$ be a $1$-form in $\Omega$. Consider the field: $$L = \omega_2 \frac{\partial}{\partial x_1} - \omega_1 \frac{\partial}{\partial x_2} \in {\frak X}(\Omega).$$I want to prove that there is $f \in {\cal C}^\infty(\Omega, \Bbb R)$, $f> 0$, with ${\rm d}(f\omega) = 0$ if and only if there is $u \in {\cal C}^\infty(\Omega, \Bbb R)$ with: $$Lu = \frac{\partial \omega_2}{\partial x_1} - \frac{\partial \omega_1}{\partial x_2}.$$
Assume $f$ exists satisfying the asked conditions. From ${\rm d}(f\omega) = 0$, I got: $$f\left( \frac{\partial \omega_2}{\partial x_1} - \frac{\partial \omega_1}{\partial x_2}\right) + Lf = 0,$$meaning I want to solve $Lu = -\frac{1}{f}Lf$, or equivalently, $Lu = f L(\frac{1}{f})$. I don't know how to do this.
For the other direction, the same calculations says that I'd want to solve that last equation for $f$, knowing $u$.
I do recognize $Lu$ as the curl of $(\omega_1,\omega_2)$, but I need something more. I also noticed that $\omega(L) = 0$ but I don't know if that'll be useful. Ideas?
Hint: Assuming you correctly derived $Lu= -\frac{1}{f}Lf$: what is $L(-\log f)$?