Galois group over $\mathbb C(t)$ 
Let $K=\mathbb C(T)$ be the rational function field over the complex field $\mathbb C$, and $L$ be the splitting field of $f(X)=X^6+2TX^3+1\in K[X]$.
(1) Find $\mathrm{Gal}(L/K)$.
(2) Find all field extensions of degree 3 containing $L/K$.

I find $L = K(\alpha)\ (\alpha =\sqrt[3]{T-\sqrt{T^2-1}} )$ and $Gal(L/K)  \cong S_3$ (Is it right?), but I can't solve (2). How to solve (2)?
 A: Write $E= K(y, z)$, where $y$ and $z$ are the roots of 
$$ X^2 + 2T X + 1.$$
Then, $E$ is a quadratic extension of $K$, and, as you implicitly point out in the comments, $yz = 1$.   
Then, as you (explicitly!) point out, the splitting field $L/K$ of the question is the splitting field over $E$ of both $X^3 -y=0$ and $X^3-z=0$: if $u^3=y$ and $uv =1$, then $v^3= z$: $L=K(u,v)=K(u)=K(v)$. 
If $\omega$ is a primitive third root of unity, the roots of $f$ (of the question) are 
$$ u,\ \omega u,\ \omega^2 u,\ v, \omega v,\text { and }\omega^2 v.$$ 
The Galois group $G$ of $L/K$ is (isomorphic to) $S_3$, and generated by 
$\sigma$ (of order three) and $\tau$ (of order 2):
$$\sigma u =\omega u \text { and } \tau u = v.$$
The elements $\sigma$ and $\tau$ are field automorphisms over $K$, and, as $L=K(u)$, are determined by their action on $u$: for instance, $$\tau v = u,\text{ and }\sigma v = \omega^2 v.$$ 
Calculating, one also finds  $$\sigma \tau \sigma^{-1} u = \omega v \text { and  } \sigma \tau \sigma^{-1} u = \omega^2 v.$$
By the Galois correspondence [the lattice of sub-groups  of $G$ and the lattice of sub-fields of $L/K$ are (anti-)isomorphic, under $H\mapsto L^H$, the fixed field of $H$], the three cubic subfields of $L$ correspond to the  fields fixed respectively by (the three sub-groups, each generated by) $\tau$, $\sigma \tau \sigma^{-1}$, and $\sigma^2\tau\sigma^{-2}$. 
(By the preceding, $\sigma$ and $\tau$ don't commute - hence the three sub-fields.)
Explicitly, if  $a = u + v \in L$, then $\tau a = a$. Calculating, one finds that if $$b = \sigma a = \omega ( u + \omega v) \text { and }c =\sigma^2 a =\omega^2(u + \omega^2 v),$$
then 
$$ ( X - a) (X-b)(X-c) = X^3 -3X +2T,$$
and the three cubic sub-extensions are $ K(a)$, $K(b)$, and $K(c)$.   
