Let's use the following numbering on the roots of your polynomial: $$\{z_1=\root3\of5,z_2=ωz_1,z_3=ω^2z_1,z_4=\sqrt3,z_5=−\sqrt3\}.$$ Any automorphism $\sigma$ of the splitting field must permute these numbers as they are the zeros of your polynomial. As the splitting field $F$ is generated by them, the automorphism $\sigma$ is fully determined once we know $\sigma(z_j),j=1,2,3,4,5.$ This gives us the usual way of identifying $\sigma$ with an element of $S_5=\operatorname{Sym}(\{z_1,z_2,z_3,z_4,z_5\})$.
But there are further constraints. The numbers $z_1,z_2,z_3$ are zeros of the factor $x^3-5$. As that factor has rational coefficients, all automorphisms must permute these three roots among themselves. Similarly for the remaining pair $z_4,z_5$ as they are the zeros of $x^2-3$. Therefore in the identification of an automorphism with a permutation in $S_5$ of the previous paragraph only the permutations in $\textrm{Sym}(\{z_1,z_2,z_3\})\times \textrm{Sym}(\{z_4,z_5\})$ are allowed. There are 12 such permutations forming a group isomorphic to $S_3\times S_2$. As you had established by other means that $[F:\mathbb{Q}]=12$, we can conclude that all such permutations come from actual automorphisms, and thus the Galois group is isomorphic to $G=S_3\times S_2$.
Let us first calculate the fixed field of the subgroup $H_1=\langle\sigma\rangle$, where $\sigma=(123)(45)$. Here we are given that $\sigma(z_1)=z_2$ and that $\sigma(z_2)=z_3$. As $\sigma$ is an automorphism of fields we get
$$
\sigma(\omega)=\sigma\left(\frac{z_2}{z_1}\right)=\frac{\sigma(z_2)}{\sigma(z_1)}=\frac{\omega^2z_1}{\omega z_1}=\omega.
$$
Therefore $\omega$ is fixed by $\sigma$, and therefore also by any power of $\sigma$.
Thus $\mathbb{Q}(\omega)\subseteq Inv(H_1)$. From Galois theory we know that $[Inv(H_1):\mathbb{Q}]=|G|/|H|=12/6=2$. As $[\mathbb{Q}(\omega):\mathbb{Q}]=2$ it follows that $\mathbb{Q}(\omega)=Inv(H_1)$.
It is fairly clear that the fixed field of the subgroup $H_2\simeq S_3$ that keeps both $z_4$ and $z_5$ fixed is $Inv(H_2)=\mathbb{Q}(\sqrt3)$.
Let's try the last subgroup $H_3$ of order 6.
I describe it as follows. We have a homomorphism $f$ from $S_3=Sym(\{z_1,z_2,z_3\})$ to
$S_2=\textrm{Sym}(\{z_4,z_5\})$ that maps a permutation $\alpha\in S_3$ to the identity element $(4)(5)$ (resp. to the 2-cycle $(45)$) according to whether $\alpha$ is an even (resp. odd) permutation. Then
$$
H_3=\{(\alpha,f(\alpha))\in S_3\times S_2\mid \alpha\in S_3\}.
$$
We easily see that $H_3$ is generated by $\beta=(123)$ and $\gamma=(12)(45)$.
Because $\beta$ acts on the set $\{z_1,z_2,z_3\}$ the same way as $\sigma$ above, we see that $\beta(\omega)=\omega$ and also that $$\beta(\sqrt{-3})=\beta(2\omega+1)=2\omega+1=\sqrt{-3}.$$ Because $\beta(z_4)=z_4$, we obviously also have $\beta(\sqrt3)=\sqrt3$. What about $\gamma$? First we get
$$
\gamma(\omega)=\gamma(\frac{z_2}{z_1})=\frac{\gamma(z_2)}{\gamma(z_1)}=\frac{z_1}{z_2}=\omega^2.
$$
As $2\omega=-1+\sqrt{-3}$ and $2\omega^2=-1-\sqrt{-3}$, this implies that $\gamma(\sqrt{-3})=-\sqrt{-3}$.
But we also have
$$
\gamma(\sqrt3)=\gamma(z_4)=z_5=-\sqrt3.
$$
Putting all these bits together we see that
$$
\gamma(i)=\gamma\left(\frac{\sqrt{-3}}{\sqrt3}\right)=\frac{-\sqrt{-3}}{-\sqrt3}=i.
$$
Similarly we see that $\beta(i)=i$. We can then conclude that $\textrm{Inv}(H_3)=\mathbb{Q}(i)$.
Somehow, the theme of symmetrization often doesn't come across very clearly in many expositions of Galois theory. Here is a basic definition:
Definition. Let $F$ be a field, and let $G$ be a finite group of automorphisms of $F$. The symmetrization function $\phi_G\colon F\to F$ associated to $G$ is defined by the formula
$$
\phi_G(x) \;=\; \sum_{g\in G} g(x).
$$
Example. Let $\mathbb{C}$ be the field of complex numbers, and let $G\leq \mathrm{Aut}(\mathbb{C})$ be the group $\{\mathrm{id},c\}$, where $\mathrm{id}$ is the identity automorphism, and $c$ is complex conjugation. Then $\phi_G\colon\mathbb{C}\to\mathbb{C}$ is defined by the formula
$$
\phi_G(z) \;=\; \mathrm{id}(z) + c(z) \;=\; z+\overline{z} \;=\; 2\,\mathrm{Re}(z).
$$
Note that the image of $\phi$ is the field of real numbers, which is precisely the fixed field of $G$. This example generalizes:
Theorem. Let $F$ be a field, let $G$ be a finite group of automorphisms of $F$, and let $\phi_G\colon F\to F$ be the associated symmetrization function. Then the image of $\phi_G$ is contained in the fixed field $F^G$. Moreover, if $F$ has characteristic zero, then $\mathrm{im}(\phi_G) = F^G$.
Of course, since $\phi_G$ isn't a homomorphism, it's not always obvious how to compute a nice set of generators for its image. However, in small examples the goal is usually just to produce a few elements of $F^G$, and then prove that they generate.
Let's apply symmetrization to the present example. You are interested in the field $\mathbb{Q}(\zeta_7)$, whose Galois group is cyclic of order $6$. There are two subgroups of the Galois group to consider:
The subgroup of order two: This is the group $\{\mathrm{id},c\}$, where $c$ is complex conjugation. You have already used your intuition to guess that $\mathbb{Q}(\zeta_7+\zeta_7^{-1})$ is the corresponding fixed field. The basic reason that this works is that $\zeta_7+\zeta_7^{-1}$ is the symmetrization of $\zeta_7$ with respect to this group.
The subgroup of order three: This is the group $\{\mathrm{id},\alpha,\alpha^2\}$, where $\alpha\colon\mathbb{Q}(\zeta_7)\to\mathbb{Q}(\zeta_7)$ is the automorphism defined by $\alpha(\zeta_7) = \zeta_7^2$. (Note that this indeed has order three, since $\alpha^3(\zeta_7) = \zeta_7^8 = \zeta_7$.) The resulting symmetrization of $\zeta_7$ is
$$
\mathrm{id}(\zeta_7) + \alpha(\zeta_7) + \alpha^2(\zeta_7) \;=\; \zeta_7 + \zeta_7^2 + \zeta_7^4.
$$
Therefore, the corresponding fixed field is presumably $\mathbb{Q}(\zeta_7 + \zeta_7^2 + \zeta_7^4)$.
All that remains is to find the minimal polynomials of $\zeta_7+\zeta_7^{-1}$ and $\zeta_7 + \zeta_7^2 + \zeta_7^4$. This is just a matter of computing powers until we find some that are linearly dependent. Using the basis $\{1,\zeta_7,\zeta_7^2,\zeta_7^3,\zeta_7^4,\zeta_7^5\}$, we have
$$
\begin{align*}
\zeta_7 + \zeta_7^{-1} \;&=\; -1 - \zeta_7^2 - \zeta_7^3 - \zeta_7^4 - \zeta_7^5 \\
(\zeta_7 + \zeta_7^{-1})^2 \;&=\; 2 + \zeta_7^2 + \zeta_7^5 \\
(\zeta_7 + \zeta_7^{-1})^3 \;&=\; -3 - 3\zeta_7^2 - 2\zeta_7^3 - 2\zeta_7^4 - 3\zeta_7^5
\end{align*}
$$
In particular, $(\zeta_7+\zeta_7^{-1})^3 + (\zeta_7+\zeta_7^{-1})^2 - 2(\zeta_7+\zeta_7^{-1}) - 1 = 0$, so the minimal polynomial for $\zeta_7+\zeta_7^{-1}$ is $x^3 + x^2 - 2x - 1$. Similarly, we find that
$$
(\zeta_7 + \zeta_7^2 + \zeta_7^4)^2 \;=\; -2 - \zeta_7 - \zeta_7^2 - \zeta_7^4
$$
so the minimal polynomial for $\zeta_7 + \zeta_7^2 + \zeta_7^4$ is $x^2+x+2$.
Best Answer
You know that $x^{3}-7$ is irreducible over $\mathbf{Q}$ so the Galois group acts transitively on the set of roots, so $\mathbf{Q}(\sqrt[3]{7})$,$\mathbf{Q}(\zeta_{3}\sqrt[3]{7})$, $\mathbf{Q}(\zeta_{3}^{2}\sqrt[3]{7})$ are subfields of degree 3. And you know that these correspond to the subgroups of order 2 of $S_{3}$, and $S_{3}$ has exactly three such subgroups. Now $\mathbf{Q}(\zeta_{3})$ is a subfield of degree 2, and hence corresponds to a subgroup of order 3 in $S_{3}$, and $S_{3}$ has a unique such subgroup.