Let $k$ be a field and $K=k(x)$ be the field of rational functions of $k(x)$. Let $\sigma $ and $\tau$ automorphisms of $K$ defined by $\sigma \left( \frac{f(x)}{g(x)} \right) = \frac{f(\frac{1}{x})}{g(\frac{1}{x})}$ and $\tau \left( \frac{f(x)}{g(x)} \right)= \frac{f(1-x)}{g(1-x)}$. Determine the fixed field $F=K^S$, where $S=\{\sigma, \tau\}$ and $Gal(K/F)$. Find an $h\in F$ such that $F=k(h)$.
The fixed field is defined as $K^S = \{x\in K \;|\; \alpha(x)=x \; \; \forall\alpha \in S\}$, so, in this case I have that an element $\frac{f(x)}{g(x)}$ belongs to the fixed field iff $\frac{f(x)}{g(x)} = \frac{f(1-x)}{g(1-x)}=\frac{f(\frac{1}{x})}{g(\frac{1}{x})}$, and then I literally don't have any clue of what to do. Any hint would be very appreciated, thanks!
$\sigma$ and $\tau$ generate a group of order six. The images of $x$ under its six elements are $$x,\quad\frac1x,\quad 1-x,\quad\frac1{1-x},\quad\frac{x-1}x,\quad\frac x{x-1}.$$ Any symmetric function of these is in the fixed field $F$. The sum of their squares is \begin{align} x^2+(1-x)^2+\frac{1+(x-1)^2}{x^2}+\frac{1+x^2}{(1-x)^2} &=\frac{H(x)}{x^2(1-x)^2} \end{align} for some degree $6$ polynomial $H$. Let $h=H(x)/(x^2(1-x)^2)$. Then $h\in F$ and $|k(x):k(h)|=6$ since $H$ has degree $6$. But $|k(x):F|$ must equal $6$, so $F=k(h)$.
Warning: this might not work in characteristic two.