$df\wedge \alpha=0$ implies $\alpha=g\,df$ for some $g$ Let $A \subset \mathbb{C}$ be an open set and $f\in {C}^{\infty}(A)$ with $df\neq0$.
I consider $\alpha$ a $1$-form such that $df\wedge \alpha=0$ and I want to prove that exists a function $g$ such that $\alpha=g\:df$.
I have that $df=f_x\,dx+{f}_y\,dy$ and $\alpha=P\,dx+Q\,dy $ for some $P,Q\in C^\infty(A) $.
Imposing
\begin{align}
0 & =df\wedge \alpha=(f_x\,dx+f_y\,dy) \wedge(P\,dx+Q\,dy) \\[6pt]
& =f_x\,dx\wedge Q\,dy+f_y\,dy\wedge P\,dx \\[6pt]
& =(f_x Q-P f_y)\,dx\wedge dy
\end{align}
I obtain $f_x Q=P f_y$. 
How can I prove now that exists a $g$ such that $P=gf_x$ and $Q=gf_y$?
Thanks for the help!

Since $df$ is nowhere $0$, you get a basis for the $1$-forms by taking $df, \star df$. (If you aren't familiar with $\star$, it's defined by $\star (a dx + b dy) = a dy - b dx$. You can check that $df\wedge\star df\ne 0$; hence the two $1$-forms are linearly independent.) So write $\alpha = s df + t{\star} df$ for some functions $s$ and $t$. Can you finish?
