if $a = ydx + xdy$. I know this is a 1 form on let's say $R^{2}$.
To evaluate $da$, we write it as $d(ydx+xdy)$ and then distribute it as $d(ydx) + d(xdy)$. Now do I apply the chain rule to get this: $dydx + ydxdx(=0) + dxdy + xdydy(=0)$
$= dydx + dxdy = 0$
or
this: $dydx + yd^{2}x + dxdy + x^{2}dy $
which one is correct?
The product rule for the wedge product (which is what you have) of a $p$-form $\alpha$ and a $q$-form $\beta$ is \begin{equation} d(\alpha\wedge\beta) =d \alpha \wedge \beta + (-1)^p\alpha\wedge d\beta. \end{equation} Note that the degree of $\beta$ actually plays no role.
The chain rule does not play a role in the case you are considering ($y$ is not a function of $x$), instead you apply the product rule, like you did in your second attempt, and the fact that $d^2=0$, so $d^2x=d^2y=0$.
So $da=0$. In fact, more holds: there is a 0-form, that is a function, $f$ such that $a = df$. Can you find $f$?
By $d^2=0$, exact forms (exterior derivative of another form) like then one you have are always closed (that is their exterior derivative vanishes). However, there are closed forms which are not exact. The prototypical example is the form \begin{equation} \frac{x dy- y dx}{x^2 +y^2} \end{equation} defined on $\mathbb{R}^2\setminus\{0\}$.