What are the intermediate fields of $\mathbb Q(\sqrt[3]2,e^{\frac{2i\pi}{3}})$ (Galois group) The elements of Galois group are
\begin{align*}
\sigma _1:\mathbb Q[\sqrt[3]2,e^{\frac{2i\pi}{3}}]&\longrightarrow \mathbb Q[\sqrt[3]2,e^{\frac{2i\pi}{3}}],\\
\sqrt[3]{2}&\longmapsto \sqrt[3]{2},\\
e^{\frac{2i\pi}{3}}&\longmapsto e^{\frac{2i\pi}{3}},
\end{align*}
\begin{align*}
\sigma _2:\mathbb Q[\sqrt[3]2,e^{\frac{2i\pi}{3}}]&\longrightarrow \mathbb Q[\sqrt[3]2,e^{\frac{2i\pi}{3}}],\\
\sqrt[3]{2}&\longmapsto \sqrt[3]{2}e^{\frac{2i\pi}{3}},\\
e^{\frac{2i\pi}{3}}&\longmapsto e^{\frac{2i\pi}{3}},
\end{align*}
\begin{align*}
\sigma _3:\mathbb Q[\sqrt[3]2,e^{\frac{2i\pi}{3}}]&\longrightarrow \mathbb Q[\sqrt[3]2,e^{\frac{2i\pi}{3}}],\\
\sqrt[3]{2}&\longmapsto \sqrt[3]{2}e^{\frac{4i\pi}{3}},\\
e^{\frac{2i\pi}{3}}&\longmapsto e^{\frac{2i\pi}{3}},
\end{align*}
\begin{align*}
\sigma _4:\mathbb Q[\sqrt[3]2,e^{\frac{2i\pi}{3}}]&\longrightarrow \mathbb Q[\sqrt[3]2,e^{\frac{2i\pi}{3}}],\\
\sqrt[3]{2}&\longmapsto \sqrt[3]{2},\\
e^{\frac{2i\pi}{3}}&\longmapsto e^{\frac{4i\pi}{3}},
\end{align*}
\begin{align*}
\sigma _5:\mathbb Q[\sqrt[3]2,e^{\frac{2i\pi}{3}}]&\longrightarrow \mathbb Q[\sqrt[3]2,e^{\frac{2i\pi}{3}}],\\
\sqrt[3]{2}&\longmapsto \sqrt[3]{2}e^{\frac{2i\pi}{3}},\\
e^{\frac{2i\pi}{3}}&\longmapsto e^{\frac{4i\pi}{3}},
\end{align*}
\begin{align*}
\sigma _6:\mathbb Q[\sqrt[3]2,e^{\frac{2i\pi}{3}}]&\longrightarrow \mathbb Q[\sqrt[3]2,e^{\frac{2i\pi}{3}}],\\
\sqrt[3]{2}&\longmapsto \sqrt[3]{2}e^{\frac{4i\pi}{3}},\\
e^{\frac{2i\pi}{3}}&\longmapsto e^{\frac{4i\pi}{3}}.
\end{align*}
We remark that
\begin{align*}
\sigma _2^3=1\\
\sigma _3^3=1\\
\sigma _4^2=1\\
\sigma _5^2=1\\
\sigma_6^2=1
\end{align*}
But $$\sigma _2\sigma _4(\sqrt[3]2)=\sigma _2(\sqrt[3]2)=\sqrt[3]2e^{\frac{2i\pi}{3}}$$
and
$$\sigma _4\sigma _2(\sqrt[3]2)=\sigma _4(\sqrt[3]2e^{\frac{2i\pi}{3}})=\sqrt[3]2e^{\frac{4i\pi}{3}},$$
therefore $\{\sigma _i\}_{i=1}^6$ is not a commutatif group and thus
$$\text{Gal}(E/\mathbb Q)=\{\sigma _i\}_{i=1}^6\cong \mathfrak S_3.$$
My problem is that it should have a $\sigma _i$ such that $\sigma _i^6=1$ and $\sigma _i^n\neq 1$ for $i\in\{1,2,3,4,5\}$ because one of my intermediate field must be $\mathbb Q$. And if not, is there a mistake in my applications $\sigma _i$?

As you have found $Aut_{\mathbb{Q}}L \simeq S_3$ then the intermediate fields of $Gal (X^3-2, \mathbb{Q})$ are in correspondence to the subgroups of $Aut_{\mathbb{Q}}L$, according to the Fundamental Theorem of Galois. Now, 
$$S_3 = \{Id, \sigma, \sigma^2, \tau, \sigma\tau, \sigma^2\tau\}$$
Notice that there is a element $\sigma$ of order $3$ generating a cyclic subgroup of order $3$, i.e., $\langle\sigma\rangle$ , and 3 elements of order $2$. What is missing is to identify such elements $\sigma$ and $\tau$ among your automorphisms $\sigma_i$ in order to find exactly the intermediate fields of $L = Gal (X^3-2,\mathbb{Q})$ through Galois correspondence.  
Hint: $\sigma = \sigma_2$ (need to check)
