There is a solution to a problem but I'm not quite sure why $[\mathbb{Q}(x) : \mathbb Q(y)] = 2$ here. Here the field is rational functions instead of the usual field extension so I don't know how to figure out $\mathbb Q(x)$ is a 2-dim $\mathbb Q(y)$-vector space.
If the function field setting is adding confusion, let $L=\mathbb{Q}(x)$ and $K=\mathbb{Q}(y)$. Then $L=K(x)$, and here, as the author shows, $x$ satisfies a quadratic equation over $K$, so $[L:K]\leq 2$. But since we know that $L/K$ is nontrivial, the degree must be exactly $2$.