Your argument up to the point where you're stuck seems completely correct to me.
To finish it, suppose that $\sigma(\omega) = \omega^j$ and that $\tau(\omega) = \omega^k$ for some $j,k \in \mathbb{N}$. Then just observe that
$$
(\sigma \circ \tau)(\omega) = \sigma(\tau(\omega)) = \sigma(\omega^k) = \sigma(\omega)^k = (\omega^j)^k = \omega^{jk}
$$
and similarly
$$
(\tau \circ \sigma)(\omega) = \tau(\sigma(\omega)) = \sigma(\omega^j) = \tau(\omega)^j = (\omega^k)^j = \omega^{jk}
$$
The Galois group of $F/\mathbb Q$ is cyclic of order $6$ as you described above, and if $\sigma$ is a generator for this group, then since the unique subgroup of order $2$ is $\{1, \sigma^2, \sigma^4 \}$, we can assume without loss of generality that $\sigma^2 = u$. If you compute $\sigma$, you can compute $\sigma^2$ and explicit $E$ as the fixed field of $\sigma^2$.
Now the extension $F/\mathbb Q$ is simple, so $\sigma$ is characterized entirely by where it maps $f$ (let me write $f = \zeta_7$ for clarity). It suffices to find a primitive root of $(\mathbb Z/7\mathbb Z)^{\times}$ and we got it. After some testing you realize that $3^6 = 1$ but $3^2, 3^3 \neq 1 \pmod 7$, so $3$ is a primitive root. This means that $\sigma(\zeta_7) = \zeta_7^3$ is a possibility for $\sigma$ as a generator of the Galois group.
Now $\sigma^2(\zeta_7) = \zeta_7^9 = \zeta_7^2$ (as promised by $\sigma^2 = u$ and $u(f) = f^2$), so we're trying to find the fixed field corresponding to the group $\{ 1,\sigma^2,\sigma^4\}$.
Because of the simple relation $\zeta_7^7 = 1$, it is not hard to solve the equation $\sigma^2(x)-x = 0$ for $x \in F$ ; write
$$
x = a_0 + a_1 \zeta_7 + \cdots + a_5 \zeta_7^5
$$
and apply $\sigma^2$ to it ; the equation $\sigma^2(x) = x$ will give you linear conditions on the coefficients. I'm afraid finding generators involves linear algebra. If you have the patience of writing down the equation by hand, you'll see the computations do not require any smart-ideas and you get
$$
E = \mathbb Q(\zeta_7 + \zeta_7^2 + \zeta_7^4).
$$
The minimal polynomial can be computed, it is
$$
(X - (\zeta_7 + \zeta_7^2 + \zeta_7^4))(X - (\zeta_7^3 + \zeta_7^5 + \zeta_7^6)) = X^2 + X + 2.
$$
The roots of this polynomial are $\frac{-1 \pm \sqrt{-7}}2$, so you can write
$$
E = \mathbb Q(\sqrt{-7}).
$$
Note : The magical reason why I get only one canonical element (i.e. $\zeta_7 + \zeta_7^2 + \zeta_7^4$) as a generator for $E/\mathbb Q$ is because $[E:\mathbb Q] = 2$, so I know the extension can only be generated by one element and that a basis for $E/\mathbb Q$ will be given by $\{1,x\}$ for some $x \in E$. This means when I tried to compute the fixed field of $\sigma^2$, I know the term for $1$ would be irrelevant and the rest would only give me one term.
Interesting remark : You know that $\mathbb Z / 6 \mathbb Z$ has only one subgroup of index $2$ and that the extension $F/\mathbb Q$ is Galois, so by the Galois correspondence $F$ has only one subfield of order $2$ over $\mathbb Q$. The element $\zeta_7 + \sigma^2(\zeta_7) + \sigma^4(\zeta_7)$ is invariant by $\sigma^2$ and is not in $\mathbb Q$, so you could've skipped the linear algebra computations and know already that $E = \mathbb Q(\zeta_7 + \zeta_7^2 + \zeta_7^4)$. We don't escape the computation of the minimal polynomial to simplify this though. (These ideas of averaging are often used in invariant theory ; I averaged the orbit of $\zeta_7$ over the subgroup $\{1,\sigma^2,\sigma^4\}$ to find the invariant substructure I wanted.)
Hope that helps,
Best Answer
Thanks to @Jyrki Lahtonen in the comments I think I understand what I missing. I will write out what I believe to be the solution. I also included the larger problem this was a part of for context.
First notice that $\sigma^3 = \tau^2 = id_{K(\omega)}$
We know that $id_{K(\omega)}$, $\sigma$, and $\sigma^2$ are linearly independent over $K(\omega)$ and so by the linear independence of field homomorphisms, we have that
$id + \omega \sigma + \omega^2 \sigma^2 \neq 0$
So there exists some $x \in K(\omega)$ such that
$x + \omega \sigma (x) + \omega^2 \sigma^2 (x) \neq 0$
Denote the expression on the LHS as $\beta$.
(1) To see that $\beta$ is not in $\mathbb{Q}(\omega)$, notice that $\sigma (\beta) = \sigma (x) + \omega \sigma^2 (x) + \omega^2 x = \omega^2 \beta \neq \beta$ and so $\beta \notin \mathbb{Q}(\omega)$
Furthermore, by computing $\beta^3$ we get the (not very elegant) equation:
$\beta^3 = x^3 + \sigma(x^3) + \sigma^2(x^3) + 3 (\omega x^2 \sigma(x) + \omega^2 x^2 \sigma^2 (x) + \omega \sigma (x^2) \sigma^2 (x) + \omega^2 x \sigma (x^2) + \omega x \sigma^2 (x^2) + \omega^2 \sigma^2(x) \sigma (x)) + 6 x \sigma (x) \sigma^2(x) $
And we can see that $\sigma(\beta^3) = \beta^3$. Thus $\beta^3 \in \mathbb{Q}(\omega)$
(2) This is simple. $\tau (\beta + \tau(\beta)) = \tau (\beta) + \tau^2(\beta) = \beta + \tau(\beta)$ and we are done.
To add some context, this was a larger part of a problem to show that with the conditions as given in the problem statement, there exists $\beta \in K(\omega)$ such that (1) and (2) hold in addition to
(3) There exists an irreducible depressed cubic $X^3 + pX + q \in Irred(\mathbb{Q})$ with $\alpha = \beta + \tau(\beta)$ as a root with splitting field $K$ and
(4) $\Delta := -27q^2 - 4p^3$ being equal to the square of some element of $\mathbb{Q}$
The cubic equation turns out to be $X^3 - 3\beta \tau(\beta)X - (\beta^3 + \tau(\beta^3))$. You can see through any discussion of Cardano's method by taking your roots to be
$\beta + \tau(\beta)$
$\omega^2 \beta + \omega \tau(\beta)$
$\omega \beta + \omega^2 \tau(\beta)$
It is a polynomial over $\mathbb{Q}$ because the coefficients are fixed by both $\sigma$ and $\tau$ (which generate the Galois group of $K(\omega) : \mathbb{Q}$). This can be checked by similar (and less tedious) calculations as before. The polynomial is irreducible because it none of its roots are in $\mathbb{Q}$ (Since $\beta + \tau(\beta) \in K$ is not fixed by $\sigma$ and so none of the roots are).
Property (4) is true because the Galois group of the polynomial (i.e. the Galois group of the extension of the splitting field of the polynomial $K : \mathbb{Q}$) is isomorphic to the alternating group $A_3$.