closed but not exact I saw several times that $\frac{-y}{x^2+y^2}dx+\frac{x}{x^2+y^2}dy$ is closed but not exact. Closed, is obvious but I can't prove non exactness, can one please help me ?
My attempt, let $f\in \omega^{0}(U)$ and $df=\frac{-y}{x^2+y^2}dx+\frac{x}{x^2+y^2}dy=\frac{\partial f}{\partial x}dx+\frac{\partial f}{\partial y} dy$ and so $\frac{\partial f}{\partial x}(x,y)=\frac{-y}{x^2+y^2},\frac{\partial f}{\partial y}(x,y)=\frac{x}{x^2+y^2}$. Now let $g(\theta)=f(\cos \theta,\sin \theta)$ so $g'(\theta)=1 \implies g(\theta)=\theta+C$. Now i'm not getting any contradiction.

You know that if you integrate over a loop into which there is no singularity, as the form is exact the integral is 0.
Then let us try with a loop containing 0 inside: let us take the circle $C(0,1)$.
$$\int_{C(0,1)} -\frac y{x^2 + y^2}dx + \frac x{x^2 + y^2}dy
= \int_0^{2\pi}  -\frac {r\sin\theta}{r^2} (-r \sin\theta d\theta)  
+ {r\cos\theta}{r^2} (r \cos\theta d\theta)\\  = 2\pi \neq 0
$$
