Let $F/K$ be a function field in one variable with genus $g$. If there is a place $P$ with degree one, then $\exists\,x,y\in F$ such that $F=K(x,y)$ and $[F:K(x)]=[F:K(y)]=2g+1$.
I'm really lost in this problem.
My first idea was to consider the case $g=1$. That would be an elliptic curve, which may be considered in the form $y^2=x^3+ax+b$ (I don't know how to prove this; I've read this is true, so I'll use it to look for ideas).
Defining $x'=x-y$, we have $y^2=(x'+y)^3+a(x'+y)+b$. Then $$[K(x',y):K(x')]=[K(x',y):K(y)]=3=2\cdot 1+1,$$
and we are done, but I still don't know what to do with the general case.
Any suggestions?
After much thinking, I came up with a much simpler solution than I expected.
If $g=0$, then $F=K(t)$ for some $t\in F$, since there is a rational place. So we just need $x=y=t$.
If $g>0$, take $s,t\in F$ such that $(s)_\infty=2gP$ and $(t)_\infty=(2g+1)P$ (this is possible because $2g,2g+1\geq 2g$). This way $[F:k(s)]=\deg(s)_\infty =2g$ and $[F:K(t)]=\deg(t)_\infty =2g+1$. Examining the degrees of the subextensions, $[F:K(s,t)]$ must divide both $2g$ and $2g+1$, which are coprime, so $[F:K(s,t)]=1\Rightarrow F=K(s,t)$.
Taking $x:=t$ and $y:=t+s$, we clearly have $F=K(x,y)$ and $(x)_\infty=(y)_\infty=(2g+1)P$, so $[F:K(x)]=[F:K(y)]=2g+1$.