The differential 1-form
$$w = \frac{x\,dy-y\,dx}{x^2+y^2}$$
is equal to the exterior derivative
$$d\theta=d(\arctan\frac{y}{x})$$
I understand by chain rule, how multivariable derivative of $\arctan(\frac{y}{x})=\frac{x-y}{x^2+y^2}$, but I have no idea how to present with $dx$ and $dy$ in the result. I do not have a good concept of what an exterior derivative is.
Note that in polar coordinates
$$\arctan\bigg( \frac{y}{x} \bigg) =\arctan\bigg( \frac{\rho \sin \theta}{\rho \cos \theta}\bigg)= \arctan (\tan\theta)=\theta$$
thus
$$d\theta=d(\arctan\frac{y}{x})$$
and
$$d\theta=\frac{\partial \theta}{\partial x}\,dx+\frac{\partial \theta}{\partial y}\,dy=\frac{-y}{x^2+y^2}\,dx+\frac{x}{x^2+y^2}\,dy=\frac{x\,dy-y\,dx}{x^2+y^2}$$