Let $dx^I$ be local coordinates and let $$\omega := \frac{x^1dx^2 - x^2dx^1}{(x^1)^2+(x^2)^2}.$$ Compute $d\omega$ for $(x^1,x^2) \ne 0$.
Remark: $d\omega$ means the exterior derivative, which we defined in the lecture as
$$d(f \ dx^I) = df \wedge dx^I$$
, where $df$ is the differential of $f$.
I can see that transfering $(x^1,x^2)$ into polar coordinates $y^I:=(r, \alpha)$ it holds $d\alpha = \omega$, so we are left to compute $d\omega = dd\alpha$. However, I do not see how to evaluate $d\alpha$.
If I am not mistaken $\alpha$ should be a $1$-form, so
$$d(\alpha \ dy^I) = d\alpha \wedge dy^I = 1 \wedge dr \wedge d\alpha,$$
but I am not sure what to do now. Could you please give me hint?
I shall write $x=x^1$ and $y=x^2\,.$ Your one-form is $$\tag{1} \omega=f\,dx+g\,dy\,,\quad f(x,y):=\frac{-y}{x^2+y^2}\,,\quad g(x,y):=\frac{x}{x^2+y^2}\,. $$ and, in Cartesian coordinates, \begin{align} d\omega&=\partial_yf\,dy\wedge dx+\partial_xg\,dx\wedge dy\\[2mm] &=\frac{(y^2-x^2)\,dy\wedge dx+(y^2-x^2)\,dx\wedge dy}{(x^2+y^2)^2}\\ &=0\,.\tag{2} \end{align} In polar coordinates, \begin{align}\tag{3} dx&=\cos\alpha\,dr-r\sin\alpha\,d\alpha\,,\quad &dy&=\sin\alpha\,dr+r\cos\alpha\,d\alpha\,,\\[2mm] dr&=\cos\alpha\,dx+\sin\alpha\,dy\,,\quad &r\,d\alpha&=-\sin\alpha\,dx+\cos\alpha\,dy\,.\tag{4} \end{align} so that indeed, from the expression for $r\,d\alpha$ in (4),
\begin{align} \omega=d\alpha\,. \end{align} This leads to \begin{align} d\omega=0 \end{align} as it should.
Another way to see that $\omega=d\alpha$ is to differentiate $$ \alpha=\arccos\Big(\frac{x}{\sqrt{x^2+y^2}}\Big) $$ w.r.t. $x$ and $y\,.$