Let $\mathbb{C}(x,y)$ be a degree $2$ extension of $\mathbb{C}(x)$ where $y$ is a root of $p(Z)=Z^2 + (x^2+1)$.
Is it true that $\mathbb{C}(x,y)$ is not a rational function field? In other words, there is no $z\in \mathbb{C}(x,y)$ such that $\mathbb{C}(x,y)=\mathbb{C}(z)$.
Geometrically
The field $\mathbb C(x,y)$ is a rational function field because it is the field of rational functions of a smooth affine conic, namely the conic $C\subset \mathbb A^2_\mathbb C$ with equation $y^2+x^2+1=0$.
Algebraically
More explicitly $\mathbb C(x,y)=\mathbb C(t)$ with $t=\frac {y-i}{x}$:
Writing $\mathbb C(x,y)=\frac{\mathbb C(x)[Y]}{(Y^2+x^2+1)}$ an explicit isomorphism is given by $$f:\mathbb C(x,y)\stackrel {\cong}{\to} \mathbb C(t)$$ where $f(x)=\frac {-2it}{1+t^2} $ and $f(y)=i\frac {1-t^2}{1+t^2} $ .
The inverse isomorphism is determined by $f^{-1}(t)=\frac {y-i}{x}$ .
The link between both approaches
We fix the point $P=(0,i)\in C$ and consider the line $y=i+tx$ through $P$ with slope $t$ (notice that then $t=\frac {y-i}{x}$ as above ).
That line cuts the conic $C$ in $P$ and in the point $P_t=(\frac {-2it}{1+t^2} ,i\frac {1-t^2}{1+t^2} )$ and this explains the formulas brutally parachuted in the previous section.