Please verify my attempted solution for the problem or correct it if necessary.
Question: Define ${\Omega}\subseteq{\mathbb{{{C}}}}$, and ${f}$ is a ${0}$-form where ${\alpha}={d}{f}\ne{0} \forall{z}\in{\Omega}$. If ${\omega}={w}{\left.{d}{z}\right.}\wedge{d}\overline{{{z}}}$ is a prescribed ${2}$-form, find ${\beta}=Pdz + Qd\overline{z}$ in ${\alpha}\wedge{\beta}={\omega}$. I should finally note that I employ less standard notation wherein $\partial_z f = \frac{\partial f}{\partial z}$ and $\partial_{\overline{z}}f=\frac{\partial f}{\partial \overline{z}}$.
Expansion of the Quantities Involved: $${\alpha}\wedge{\beta}={d}{f}\wedge{\beta}={\left(\partial_{{{z}}}{f}{\left.{d}{z}\right.}+\partial_{{\overline{{{z}}}}}{f}{d}\overline{{{z}}}\right)}\wedge{\beta}={\omega}={w}{\left.{d}{z}\right.}\wedge{d}\overline{{{z}}}$$ $${\alpha}\wedge{\beta}={\det{{\left(\begin{array}{cc} \partial_{{{z}}}{f}&\partial_{{\overline{{{z}}}}}{f}\\{P}&{Q}\end{array}\right)}}}{\left.{d}{z}\right.}\wedge{d}\overline{{{z}}}$$ $${\alpha}\wedge{\beta}=\partial_{{{z}}}{f}{Q}-\partial_{{\overline{{{z}}}}}{f}{P}{\left.{d}{z}\right.}\wedge{d}\overline{{{z}}}$$
Verification Needed: Would ${Q}{\left({z}\right)}={w}{\left(\partial_{{{z}}}{f}\right)}^{{-{1}}}$ and ${P}={0}$ suffice?