I'm trying to solve the following problem, but it's too difficult for me.
Let $\mathbb{C}(T)$ be the rational function field over the complex field $\mathbb C$, and put $L:=\mathbb{C}(T,\sqrt{T^2 + T +1})$, $K:=\mathbb{C}(T^3)$. Let $M$ be a Galois closure of $L$ over $K$.
(1) Calculate $[M:K]$.
(2) Find all intermediate fields of $L$ and $K$.
What should I do first?
This is a very instructive example. Let me explain in detail (certainly not in the most efficient way, but hopefully it will be enlightening) how we can work out the Galois group in this situation, which is a bit more than what was asked for.
Let $K = \mathbb{C}(t^3)$ and $L = \mathbb{C}(t,\sqrt{t^2+t+1})$ as you defined (I think the use of a lowercase $t$ for the indeterminate makes things easier to read), and let $E = \mathbb{C}(t)$. Consider all these fields embedded in a common algebraic closure $\Omega$ (which we can think of as a subfield of Puiseux series in $t-a$ for some $a$ or in $t^{-1}$). For convenience, let $\omega := e^{2i\pi/3}$. Thus, $t^2 + t + 1 = (\omega t-1)(\omega^2 t-1)$. We will also make frequent use of the polynomials $\omega^2 t^2 + \omega t + 1 = (t-1)(\omega^2 t-1)$ and $\omega t^2 + \omega^2 t + 1 = (t-1)(\omega t-1)$.
First note that $E$ is Galois over $K$ with Galois group $\mathbb{Z}/3\mathbb{Z}$ acting on $t$ by taking it to $\omega^i t$ for $i \in \mathbb{Z}/3\mathbb{Z}$. Also, $L$ is Galois over $E$ with Galois group $\mathbb{Z}/2\mathbb{Z}$ acting on $\xi_0 := \sqrt{t^2+t+1}$ by taking it to $(-1)^j \sqrt{t^2+t+1}$ for $j \in \mathbb{Z}/2\mathbb{Z}$.
The conjugates of $\xi_0 := \sqrt{t^2+t+1}$ over $K$ are obviously among $\pm\xi_0,\pm\xi_1,\pm\xi_2$ where $\xi_1 := \sqrt{\omega^2 t^2 + \omega t + 1}$ and $\xi_2 := \sqrt{\omega t^2 + \omega^2 t + 1}$ (for the moment, these square roots are chosen arbitrarily in $\Omega$, but we will correct this shortly). Indeed, any automorphism of $\Omega$ over $K$ has to take $\xi_0^2$ to some $\xi_i^2$ so $\xi_0$ to some $\pm\xi_i^2$. (As a matter of fact, we can check that the polynomial $$(X^2 - (t^2+t+1)) (X^2 - (\omega^2 t^2 + \omega t + 1)) (X^2 - (\omega t^2 + \omega^2 t + 1)) = X^6 - 3 X^4 - 3(u-1) X^2 - (u^2 - 2u + 1)$$ — where we let $u=t^3$, is irreductible over $K = \mathbb{C}(u)$, so that the $\pm\xi_0,\pm\xi_1,\pm\xi_2$ are exactly the conjugates of $\xi_0$. But we won't need this now and we'll prove something stronger anyway.)
So now the Galois group $G$ of the Galois closure $M$ of $L$ over $K$ will be seen as a permutation group over $\pm\xi_0,\pm\xi_1,\pm\xi_2$ and the question is, which one. We know that $G$ surjects onto $\mathbb{Z}/3\mathbb{Z} = \mathrm{Gal}(E/K)$ by acting on the partition $\{\{\pm\xi_0\},\{\pm\xi_1\},\{\pm\xi_2\}\}$. (Also the parenthesis in the last paragraph means exactly that $G$ acts transitively on these six elements, but this is not sufficient to determine it, and as I said we'll work it out differently.) Clearly, $G$ is a subgroup of the group $\mathfrak{G}$ of permutations of $\pm\xi_0,\pm\xi_1,\pm\xi_2$ that are compatible with the partition $\{\{\pm\xi_0\},\{\pm\xi_1\},\{\pm\xi_2\}\}$: this $\mathfrak{G}$ is the wreath product $(\mathbb{Z}/2\mathbb{Z}) \wr (\mathbb{Z}/3\mathbb{Z})$, i.e. the semidirect product $(\mathbb{Z}/2\mathbb{Z})^3 \rtimes (\mathbb{Z}/3\mathbb{Z})$ with the obvious action, and it has order $24$. One might naïvely think that $G$ will be equal to $\mathfrak{G}$, but it turns out to be smaller. So we need to find some relation to show this.
Now notice this: $$(t^2+t+1) (\omega^2 t^2 + \omega t + 1) (\omega t^2 + \omega^2 t + 1) = (t-1)^2 (\omega t-1)^2 (\omega^2 t-1)^2 = (t^3-1)^2$$
In other words, the square of $\xi_0 \xi_1 \xi_2$ is $(t^3 - 1)^2$, so that $\xi_0 \xi_1 \xi_2$ is plus or minus $t^3 - 1$, which belongs to $K$. Let us set $\xi_0 \xi_1 \xi_2 = t^3 - 1$ by changing the sign of $\xi_2$ if necessary. Since this quantity belongs to $K$, it must be fixed by any element of $G$. This means that, in fact, Galois cannot change the sign of all three $\xi_i$ independently, but only of two of them, and the third is then determined. In other words, $G$ is contained in the subgroup of $\mathfrak{G}$ (of index $2$, order $12$) consisting of those permutations which send $(\xi_0,\xi_1,\xi_2)$ to $(\varepsilon_0 \xi_i, \varepsilon_1 \xi_{i+1}, \varepsilon_2 \xi_{i+2})$ with $\varepsilon_0 \varepsilon_1 \varepsilon_2 = +1$ (and indices are mod $3$, of course). Equivalently, the Galois closure $M$ of $L$ over $K$ is contained in $E(\xi_0,\xi_1)$ (since $\xi_2$ belongs to this), which has degree $4$ over $E$ (and $12$ over $K$).
Now it remains to explain why the Galois group (and $M$) is no smaller than this.
But $M$ is contained in the field $N = \mathbb{C}(\sqrt{t-1},\sqrt{\omega t-1},\sqrt{\omega^2 t-1})$ by viewing $\xi_0$ as $\sqrt{t-1} \cdot \sqrt{\omega t-1}$ and so on (this is coherent with our prescription of signs $\xi_0 \xi_1 \xi_2 = t^3-1$). Now $N$ is Galois over $E$ with Galois group $(\mathbb{Z}/2\mathbb{Z})^3$ acting independently on the signs of $\sqrt{t-1},\sqrt{\omega t-1},\sqrt{\omega^2 t-1}$: this is well known and can be checked, for example, by embedding everything in the algebraically closed field $\mathbb{C}(((t-a)^{1/\infty}))$ of Puiseux series in $t-a$ for $a = 1,\omega^2,\omega$, which gives an automorphism taking $\sqrt{t-a}$ to $-\sqrt{t-a}$ and fixing every $\sqrt{t-b}$ for $b\neq a$. Clearly, the automorphism of $N$ taking $\sqrt{t-1}$ to $-\sqrt{t-1}$ and fixing $\sqrt{\omega t-1}$ and $\sqrt{\omega^2 t-1}$ takes $\xi_0,\xi_1,\xi_2$ to $+\xi_0,-\xi_1,-\xi_2$ respectively. In this way, we see that $G$ contains every permutation that flips the sign of an even number of the $\xi_i$ (without otherwise permuting them). And since we also know that $G$ must surject onto $\mathbb{Z}/3\mathbb{Z}$ acting on the partition $\{\{\pm\xi_0\},\{\pm\xi_1\},\{\pm\xi_2\}\}$, it is clear that $G$ is exactly the group of order $12$ announced below.
Conclusions:
$G$ is the group of order $12$ described as $S \rtimes (\mathbb{Z}/3\mathbb{Z})$ where $S$ is the subgroup of index $2$ of $(\mathbb{Z}/2\mathbb{Z})^3$ consisting of elements having zero sum (viz., $\{(0,0,0), (0,1,1), (1,0,1), (1,1,0)\}$), and acting on $\pm\xi_0,\pm\xi_1,\pm\xi_2$ by having $S$ act by an even number of sign changes and $\mathbb{Z}/3\mathbb{Z}$ act by cyclic permutations of $\xi_0,\xi_1,\xi_2$.
In particular, the Galois closure $M$ of $L$ over $K$ has degree $\#G = 12$ over $K$ (note that $M$ is generated over $E$, and even over $K$, by any two of $\xi_0,\xi_1,\xi_2$).
From this, it is easy to work out intermediate fields: there are exactly $10$ intermediate fields between $K$ and $M$ corresponding to the $10$ easily found subgroups of $G$. (If we want fields between $K$ and $L$, we want subgroups of $G$ containing the order $2$ subgroup $\mathrm{Gal}(M/L)$ whose unique non-trivial element changes the signs of $\xi_1$ and $\xi_2$ simultaneously: it's fairly obvious that there are none other than $G = \mathrm{Gal}(M/K)$ and $\mathrm{Gal}(M/E)$ and $\mathrm{Gal}(M/L)$, so $K,E,L$ are the only intermediate fields between $K$ and $L$.)
[The most efficient approach is probably this: immediately define $N = \mathbb{C}(\sqrt{t-1},\sqrt{\omega t-1},\sqrt{\omega^2 t-1})$, note that it is Galois not only over $E$ but also over $K$, compute its Galois group (which, as an abstract group, is isomorphic to $\mathfrak{G}$ above), and note that the only element of the latter which fixes $M$ is the one which changes the sign of each of $\sqrt{t-1},\sqrt{\omega t-1},\sqrt{\omega^2 t-1}$.]