Start with $R=\mathbb{C}(u,v)$ and consider its quadratic extension $L=R(\sqrt{u})=R(s)=\mathbb{C}(s,v)$ with $s^2=u$.  

Take an element $z$ of $L$, viewed as a rational function $z(s,v)$. Then $R(z)=L$ iff $z\notin R$ (because $[L:R]=2$) iff $\overline{z}\neq z$ where $\overline{z}$ is the conjugate of $z$ which is $z(-s,v)$.  

We can also view $L$ as $\mathbb{C}(x,y)$ in many different ways; let us pick 
$$x=s+v, \quad y=s+2v.$$
For $i, j\in \mathbb{N}$ we have $x^i y^j=
(s+v)^i (s+2v)^j$ and $\overline {x^i y^j}=
(-s+v)^i (-s+2v)^j$. If $(i,j)\neq(0,0)$ these are always different, so $x^i y^j\notin R$ and thus $R(x^i y^j)=L$.