Question on Differential forms Let $\Omega\subseteq \mathbb{R}^2$ be an open set and $\omega=\omega^1dx_1+\omega^2dx_2$ an $1$-form on $\Omega$ and 
$$L=\omega^2\frac{\partial}{\partial x_1}-\omega_1\frac{\partial}{\partial x_2}\in\mathfrak{X}(\Omega)$$
How to prove that exist $f\in C^{\infty}(\Omega)$, such that $f>0$ and $d(f\omega)=0$, iff there exists an $u\in C^{\infty}(\Omega)$, such that 
$$Lu=\frac{\partial\omega^2}{\partial x_1}-\frac{\partial\omega^1}{\partial x_2}$$
 A: The standard result is that if $d(f\omega) = 0$, with $f\ne 0$, then $d\omega\wedge\omega = 0$. The converse follows (locally, at least) from the Frobenius Theorem on differential systems/integral submanifolds. 
First I'm going to assume $N=2$.
Since $-(Lu)\, dx_1\wedge dx_2 = \omega\wedge du$ by definition, your criterion $-(Lu)\,dx_1\wedge dx_2=d\omega$ can be rephrased as
$$\omega\wedge du = d\omega\,.$$
But
$$d(f\omega) = f\,d\omega + df\wedge\omega = 0 \iff d\omega = -\frac{df}f\wedge\omega
\iff d\omega = \omega\wedge \frac{df}f\,,$$
so taking $du = df/f$, i.e., $u=\log f$ does it. Conversely, given the $u$, we take $f=e^u$ and $d(f\omega)=0$.
Now, in general, with $N>2$, we add a "$\wedge dx_3\wedge\dots\wedge dx_N$" to the above computations. It is certainly still true that if $d(f\omega)=0$, then there exists $u$.  But the converse doesn't seem to work: With the hypothesis on $Lu$, we'll only get $d(f\omega)=0 \pmod{\langle dx_3,\dots,dx_N\rangle}$. Counterexample forthcoming?
Edit: Ah, I just saw you amended it to be $N=2$, so I won't construct a counterexample for $N=3$ :)
