Let $K=F(t)$ with $t$ transcendental over F meaning that $K$ is a simple transcendental extension of F.
Let $\sigma_1, \sigma_2 \in \Gamma(K:F)$ the Galois group of K:F.
$\sigma_1 : t \mapsto t^{-1}$ $\sigma_2 : t \mapsto 1-t$
It is obvious that $|\sigma_1|=|\sigma_2|=2$ but I want to know how to find the $Fix(<\sigma_1>)$ and $Fix(<\sigma_2>)$ I know that F(u)=$Fix(<\sigma_1>)$ must satisfy $|K:F(u)|=2$ and I have an answer as $ u_1= t+t^{-1}$ for $\sigma_1$ and $u_2=t(1-t)$ for $\sigma_2$ but I have no idea how these are found or if it's just trial and error.
If $\sigma$ is an automorphism of degree $n$, then $\sigma$ permutes $\{t_i:=\sigma^i(t)\mid 0\leqslant i<n\}$, so it fixes all elementary symmetric polynomials $s_i:=s_i(t_0,\ldots,t_{n-1})$, $1\leqslant i\leqslant n$. Thus $\sigma$ fixes $L:= F(s_1,\ldots,s_n)$, i.e. $L\leqslant Fix(\langle\sigma\rangle)$. On the other hand, $t$ satisfies polynomial $(X-t_0)(X-t_1)\ldots (X-t_{n-1})$, which has coefficients in $L$ (its coefficients are exactly $s_i$'s), so $|K:L|\leqslant n$. But $|K:Fix(\langle \sigma\rangle)|=n$ and $L\leqslant Fix(\langle\sigma\rangle)$, by chain rule we conclude $L=Fix(\langle\sigma\rangle)$.
Therefore, $Fix(\langle \sigma\rangle)$ is determined by elementary symmetric polynomials, so you should calculate them. In your case, $n=2$, they are just sum and product. $Fix(\langle\sigma_1\rangle)= F(t+\sigma_1(t),t\cdot\sigma_1(t))= F(t+t^{-1},t\cdot t^{-1})= F(t+t^{-1})$, and similarly for $\sigma_2$.