Why is $\cfrac{y \text{d}x - x \text{d}y}{x^2+y^2}$ closed?
I have calculated $\text{d}\left(\cfrac{y \text{d}x - x \text{d}y}{x^2+y^2}\right) = -2 \cfrac{(x \text{d}y - y\text{d} x)(x\text{d}x+ y\text{d}y)}{(x^2+y^2)^2}$.
And what's next? How to get $\text{d}\left(\cfrac{y \text{d}x - x \text{d}y}{x^2+y^2}\right) = 0$?
It's not clear how you performed your calculation, but I will show you how such a computation should go. If we call $$\omega=\frac{y\ dx-x\ dx}{x^2+y^2},$$ then
\begin{align*} d\omega&=d\left(\frac{y}{x^2+y^2}\right)\wedge dx+d\left(-\frac{x}{x^2+y^2}\right)\wedge dy\\ &=\frac{(x^2+y^2)dy-y(2x dx+2y dy)}{(x^2+y^2)^4}\wedge dx-\frac{(x^2+y^2)dx-x(2x dx+2ydy)}{(x^2+y^2)^4}\wedge dy\\ &=\frac{(x^2+y^2)dy-2y^2 dy}{(x^2+y^2)^4}\wedge dx-\frac{(x^2+y^2)dx-2x^2 dx}{(x^2+y^2)^4}\wedge dy\\ &=\frac{y^2-x^2}{(x^2+y^2)^2}dx\wedge dy+\frac{x^2-y^2}{(x^2+y^2)^2}dx\wedge dy\\ &=\frac{y^2-x^2+x^2-y^2}{(x^2+y^2)^2}dx\wedge dy\\ &=\boxed{0.} \end{align*}