I get the result from this post Prove that there are infinitely many intermediate field of $K=k(x,y)$ over $F=k(x^p,y^p)$.
Assume we already have result
If $\theta\in K\setminus F$, then $[F(\theta): F]=p$
Let $h(t)=t-\theta\in F(\theta)[t]$. It has root in $K$ and clearly irreducible, hence $[K:F(\theta)]=\deg(h)=1$ and hence $[L:F]=[L:F(\theta)][F(\theta):F]=p$. I don't really see where I went wrong in the proof. Can anyone point it out? Thanks in advance.
To prove $[k(x,y):k(x^p,y^p)]=p^2$, we can argue as follows . . .
Consider the chain of fields $$ k(x^p,y^p) \subset k(x^p,y) \subset k(x,y) \qquad\qquad\;\;\; $$ Since $p$ is prime, and the inclusions are proper, it follows that $$ \begin{cases} [k(x,y):k(x^p,y)]=p \qquad\qquad\qquad\;\;\;\;\;\, \\[4pt] [k(x^p,y):k(x^p,y^p)]=p\\ \end{cases} $$ hence \begin{align*} &[k(x,y):k(x^p,y^p)]\\[4pt] &=[k(x,y):k(x^p,y)]\cdot[k(x^p,y):k(x^p,y^p)]\\[4pt] &=p\cdot p\\[4pt] &=p^2\\[4pt] \end{align*} As regards your error . . .
You don't know that $F(\theta)=K$, so the fact that $\theta$ has degree $1$ over $F(\theta)$ doesn't prove $[K:F(\theta)]=1$.
Thus, from the fact that $\theta \in K$, and $t-\theta$ is irreducible in $F(\theta)[t]$, all you get is $[F(\theta):F(\theta)]=1$, which tells you nothing about $[K:F]$.