Consider the vector field $X = x \frac{\partial }{\partial x} + y \frac{\partial }{\partial y} $ on $\mathbb{R^2}$. I need to find a $1$-form $\omega$ on $\mathbb{R^2} - \{ (0,0)\}$ such that $\omega(X) = 0$.
Here's what I tried:
$$\omega(X) = (a \ dx + b \ dy) \Big( x \frac{\partial }{\partial x} + y \frac{\partial }{\partial y}\Big) = ax + by = 0.$$
Now, since $a= -y, b=x$ is the solution of above equation, we get the following $1$-form
$$\omega = -y \ dx + x \ dy.$$
But since I need to find $\omega$ on $\mathbb{R^2} - \{ (0,0)\}$ so there are several other values of $a$ and $b$ possible. Like $a= x/y$, $b = -x^2 / y^2 $. So which one I need to consider here?
The one-form $\omega = -y dx + xdy$ on $\mathbb{R}^2\setminus\{(0, 0)\}$ satisfies the condition $\omega(X) = 0$ as would any multiple of $\omega$.
However, the other form you suggested, $\frac{x}{y}dx - \frac{x^2}{y^2}dy$ doesn't work as it is not defined on $\mathbb{R}^2\setminus\{(0, 0)\}$ (it is undefined whenever $y = 0$).