Let $K=\mathbb{C}(x)$ where $x$ is transcendental over $\mathbb{C}$. Let $\sigma$ be the automorphism of $K$ over $\mathbb{C}$ given by $\sigma(x)=\zeta x$ where $\zeta$ is a primitive 3rd root of unity. Let $\tau$ be the automorphism given by $\tau(x)=x^{-1}$. We have $\sigma^3=1=\tau^2$ and $\tau\sigma=\sigma^{-1}\tau$. Then group $G$ generated by $\sigma$ and $\tau$ is isomorphic to $D_6$ which has order 6.
I'm trying to show the fixed field of $G$ is $\mathbb{C}(x^3+x^{-3})$ and was stuck. Also do we still have $[\mathbb{C}(x):\mathbb{C}(x^3+x^{-3})]=6=|G|?$
Write $t=x^3+x^{-3}$. It is easy to check that $\Bbb{C}(t)$ is contained in the fixed field $\mathrm{Inv}(G)$ of $G$.
By Galois theory we know that $[\Bbb{C}(x):\mathrm{Inv}(G)]=6$. Therefore it suffices to show that $[\Bbb{C}(x):\Bbb{C}(t)]\le6$. This is because the tower law tells us that $$ [\Bbb{C}(x):\Bbb{C}(t)]=[\Bbb{C}(x):\mathrm{Inv}(G)]\cdot[\mathrm{Inv}(G):\Bbb{C}(t)], $$ when we can conclude that the latter factor is $\le1$ giving us the claim $$\mathrm{Inv}(G))=\Bbb{C}(t).$$ But, as commented, $x$ is a zero of the polynomial $$ P(T)=T^6-tT^3+1\in\Bbb{C}(t)[T]. $$ Therefore $[\Bbb{C}(t)(x):\Bbb{C}(t)]\le\deg P(T)=6$. Clearly $\Bbb{C}(t)(x)=\Bbb{C}(x)$, so we are done.