Is the subextension of a purely transcendental extension purely transcendental over the base field?

This is a difficult question known as the "Lüroth problem", and the answer depends on the transcendence degree of $K/F$. In the transcendence degree $1$ case, any subextension must be purely transcendental, and this is known as Lüroth's theorem. For transcendence degree $2$ or higher, it is possible for a subextension to not be purely transcendental. In the special case that the base field $F$ is algebraically closed and of characteristic $0$, any subextension must still be purely transcendental in the transcendence degree $2$ case, but there are counterexamples for transcendence degree $3$. In all cases except the transcendence degree $1$ case, the proofs of these statements are quite hard and use some heavy machinery from algebraic geometry (by thinking of the field $E$ as the function field of a variety over the base field $F$).

The answer is no in general. This type of questions are related to birational geometry.

See https://en.wikipedia.org/wiki/Rational_variety

Where it links to a paper by Swan, which shows that

Let $G\simeq\mathbb Z/p\mathbb Z$ acts transitively on $x_1, \cdots, x_p$. Then $\mathbb Q(x_1, \cdots, x_p)^G$ is not purely transcendental over $\mathbb Q$ for $p=47$.

I haven't read this paper, and I don't know whether there is an easier construction.