If $F$ is any algebraically closed field, and $L \supset F(X)$ is a finite extension of the purely transcendental extension of $F$ of transcendence degree $1$, then can $L$ necessarily be embedded into $F(Y)$ (so that $X$ becomes a rational function in $Y$)?
It seems like this ought to be an easy question to google, but I haven't found anything. I did find this wikipedia article which indicates a connection to algebraic geometry, so if my grasp of algebraic geometry were strong enough, I would probably be able to answer this question on my own. Alas, I am not there yet. :)
A field extension in this situation corresponds to a ("ramified") cover of an algebraic curve by another. $F(X)$ is the function field of the projective line. Genus is a useful invariant of these things: it can only increase (or not decrease, to be precise) in an extension, by the Riemann-Hurwitz relation. So, yes, there are some extension that do not increase genus (like $Y^2=X$), but in general the genus goes up. So the extension cannot be written as $F(Y)$ for any $Y$... or else it'd be the projective line, of genus $0$.
But, also, this is not quite to say that there's no $Y$ in the extension so that $X$ is a rational function of it. But maybe that's not what you intended to ask.