Let $k$ be a field, and let $K=k(x)$ be the rational function field in one variable over $k$. Let $\sigma$ and $\tau$ be the automorphism of $K$ defined by $\sigma(\frac{f(x)}{g(x)})=\frac{f(\frac{1}{x})}{g(\frac{1}{x})}$ and $\tau(\frac{f(x)}{g(x)})=\frac{f(1-x)}{g(1-x)}$, respectively. Determine the fixed field $F$ of $\{σ,τ\}$.
To calculate $F$, I let $\frac{f(x)}{g(x)}=\frac{f(\frac{1}{x})}{g(\frac{1}{x})}$ and $\frac{f(x)}{g(x)}=\frac{f(1-x)}{g(1-x)}$, but I can't compute a rational function that satisfies both $\frac{f(x)}{g(x)}=\frac{f(\frac{1}{x})}{g(\frac{1}{x})}$ and $\frac{f(x)}{g(x)}=\frac{f(1-x)}{g(1-x)}$
Let us give a lemma: the finite Group $G$ act on $K=k(x)$, then $f\mapsto\frac{1}{|G|}\sum\limits_{\phi\in G}\phi(f)$ gives a surjection of $K\to K^{G}$.
By this lemma,$$K^{A_3}=Im(id+\tau\sigma+(\tau\sigma)^2)$$Let $t_1=x$, $t_2=\tau\sigma(x)=\frac{1}{1-x}$, $t_3=(\tau\sigma)^2(x)=1-\frac{1}{x}$, then $$K^{A_3}=\{f(t_1)+f(t_2)+f(t_3):f\in K\}$$Let $$s_1=t_1+t_2+t_3=1+\frac{1}{1-x}-\frac{1}{x}+x,$$ $$s_2=t_1t_2+t_1t_3+t_2t_3=-2+\frac{1}{1-x}-\frac{1}{x}+x,$$ $$s_3=t_1t_2t_3=-1,$$then $K^{A_3}=k(s_1, s_2, s_3)$. Let $\xi = \frac{1}{1-x}-\frac{1}{x}+x$, then $K^{A_3}=k(\xi)$.
By the same way, we can calculate $(K^{A_3})^\sigma = k(h)$, which $h=x+\frac{1}{x}-x^2-\frac{x}{(1-x)^2}$. In the fact, $F=(K^{A_3})^\sigma=K^{S_3}$. So $F=k(h)$ and $Gal(K/F)=S_3$.