ac.commutative algebra – How to show an invariant subfield of rational function field $mathbb{Q}(x)$ under a certain group action is actually a simple extension?

Let $K=mathbb{Q}(x)$ be the rational functions in one variable $x$ and let the automorphisms $phi,psi$ of $K$ be defined as $phi(x)=-frac{1}{x+1}$ and $psi(x)=frac{1}{x}$.

Let $G$ be the group generated by $phi,psi$, then $G=langle phi,psi|phi^3=psi^2=1,phipsi=psiphi^2rangle $.

To be specific, $ G={1,phi,phi^2,psi,phipsi,psiphi} $ and
$$ phi(x)=-frac{1}{x+1}\ phi^2(x)=-frac{x+1}{x}\ psi(x)=frac{1}{x}\ phipsi(x)=-x-1\ psiphi(x)=-frac{x}{x+1} $$
let $K_0$ be the invariant subfield of $K$ under the $G$-action.

The point is to show that $K_0$ is a simple extension of $mathbb{Q}$.

The threads I hold: Regard $K$ as a $G$-module then $K_0$ is nothing but the 0-th cohomology of $G$ with coefficient $K$ and $text{Gal}(K/K_0)=G$. On the other hand set $N=sumlimits_{gin G}g$ to be the norm element of $G$, then there is a cannonical map $alpha$ form $K$ to $K_0$ sending every rational function $fin K$ to $Nf$. And I guess $alpha$ is a surjection which I’m not sure. Notice that $Nx=-3$ and $N(x^2)=(t+1)^2+(frac{1}{t+1})^2+t^2+frac{1}{t^2}+(frac{t+1}{t})^2+(frac{t}{t+1})^2$ and I have a vague sense that $N(x^i)$ could be expressed in $N(x^2)$ for any $iinmathbb{Z}$. So I guess $K_0=mathbb{Q}(N(x^2))$. Again, I’m not sure about this.