example for calculation of the exterior derivative I don't understand how to calculate exterior derivatives. For the form
$$
    \theta = \frac{x\, dy - y\, dx}{x^2 + y^2}
$$
I arrive at the following solution:
$$
\begin{align*}
    d\theta
    &= d\left(\frac{x}{x^2 + y^2}\right) \wedge dy
     + d\left(\frac{-y}{x^2 + y^2}\right) \wedge dx \\
    &= \frac{-x^2 + y^2}{(x^2 + y^2)^2} dx \wedge dy
     - \frac{x^2 - y^2}{(x^2 + y^2)^2}dy \wedge dx \\
    &= 0.
\end{align*}
$$
I don't understand the second step. Or also in these examples Exterior Derivatives
I don't see how we get the solutions.
I know that $df_p$ is a function from $T_pM$ to $\mathbb{R}$.
But here we don't have a specific point $p$ and I don't think I can use this definition to calculate the exterior derivative of $x/(x^2 + y^2)$. Is there another definition for $df$?
I was also wondering if there are general rules that could help with this kind of calculations (such as $dx \wedge dx = 0$).

If you apply the definition you have:
$$
\begin{align*}
    d\left(\frac{x}{x^2 + y^2}\right) \wedge dy
    &= \partial_x\left(\frac{x}{x^2 + y^2}\right) dx \wedge dy
     + \partial_y\left(\frac{x}{x^2 + y^2}\right) dy \wedge dy \\
    &= \partial_x\left(\frac{x}{x^2 + y^2}\right) dx \wedge dy \\
    &= \frac{-x^2 + y^2}{(x^2 + y^2)^2} dx \wedge dy
\end{align*}
$$
I'll let you conclude the exercise.
