every one-form on $\mathbb{R}^2$ is locally $f\,dg$ I have read (McDuff-Salamon, Introduction to Symplectic Topology) that every smooth one-form on $\mathbb{R}^2$ can be locally written as $f\, dg$ for $f, g$ smooth functions. To my embarrassment I don't seem to be able to prove this. Can anyone help me out?
Some ideas:
This is equivalent to saying that for all functions $\alpha, \beta$ on $\mathbb{R}^2$, there locally exist solutions to
\begin{align}
f(x, y) \partial_x g(x, y) &= \alpha(x, y) \\
f(x, y) \partial_y g(x, y) &= \beta(x, y)
\end{align}
So it seems that it is necessary to be able to solve the PDE
$$
\frac{\partial_x g}{\partial_y g} = \gamma
$$
for arbitrary functions $\gamma$. But we also need to worry about the case where $\beta$ is zero at some points...
 A: The precise claim is that if $\omega$ is non-zero at $p \in \mathbb{R}^2$ then $\omega$ can be written locally around $p$ as $\omega = f dg$. To see that this holds, choose some non-zero vector field $X$ such that $\omega(X) = 0$ locally around $p$. Since $X \neq 0$, one can find a coordinate system $(x,y)$ around $p$ with $X = \frac{\partial}{\partial x}$ (i.e "flatten the vector field"). Writing $\omega = \alpha dx + \beta dy$ we see that $0 = \omega(X) = \alpha$ so around $p$ we have $\alpha \equiv 0$ and hence $\omega = \beta dy$.

To see that this is not always possible when $\omega|_{p} = 0$, consider the form $\omega = -y dx + x dy$ around the origin. If we can write
$$ -y dx + x dy = fdg = f(x,y)g_x(x,y) dx + f(x,y) g_y(x,y) dy $$
then we get
$$ -y = f(x,y) g_x(x,y), \,\,\, x = f(x,y) g_y(x,y). $$
By plugging in $(x,0)$ in the first equality, we get $f(x,0) g_x(x,0) = 0$ and by plugging in $(0,y)$ in the second equality, we get $f(0,y) g_y(0,y) = 0$. Since $f(x,y) \neq 0$ for all $(x,y) \neq 0$ we get that $g_x(x,0) = 0$ for all $x \neq 0$ and $g_y(0,y) = 0$ for all $y \neq 0$. By continuity, we must have $g_x(0,0) = g_y(0,0) = 0$ (or $dg|_{(0,0)} = 0$). Since
$$ d\omega = 2(dx \wedge dy) = df \wedge dg $$
and $dg|_{(0,0)} = 0$ while $d\omega|_{(0,0)} \neq 0$, we get a contradiction.
Away from the origin, we can always write $\omega = r^2 d\theta$.
