As the title says, I would like to find an example for a cyclic field extension $K/\mathbb{Q}_5$ with $[K:\mathbb{Q}_5]=6$ and $e(K/\mathbb{Q}_5)=3$.
Because this implies that the intertial degree of $K/\mathbb{Q}_5$ must be $2$, the extension $K_0 = \mathbb{Q}_5(\zeta_3)/\mathbb{Q}_5$ (where $\zeta_3$ is a primitive third root of unity) is unramified of degree $2$.
Now I only need to find an element $\alpha$ such that $K=K_0(\alpha)/K_0$ is totally ramified of degree $3$. As this extension must be cyclic (because it has prime degree), the element $\alpha$ must satisfy $\alpha,\alpha^2 \not\in L_0$ but $\alpha_3 \in L_0$. Or one could say that the polynomial $x^3 - \alpha^3$ is irreducible over $L_0[x]$.
I wanted to give $\alpha = \sqrt[3]{2}$ a try (i.e. $\alpha^3 = 2$) and noted that $x^3-2$ has the root $3$ over $\mathbb{F}_5$. This only says that if $\sqrt[3]{2} \in K$, then $K/K_0$ is indeed totally ramified. But this still says nothing about whether $x^3-2$ is irreducible over $K_0[x]$ or not.
Could you please guide me finishing my example?
The result alluded to in reuns' comment can be found in Lang's Algebraic Number Theory, Proposition II.5.12: Let $K$ be a complete, discretely valued field with perfect residue field, and let $p \geq 0$ be the characteristic of the residue field. Let $E/K$ be a field extension that is totally ramified (meaning the degree $[E : K]$ is equal to the ramification index $e$) and tamely ramified (meaning $e$ is not divisible by $p$). Then there exist uniformizers $\pi \in K$ and $\Pi \in E$ satisfying $\Pi^e = \pi$, and so $E \cong K(\pi^{1/e})$.
Thus, if $K$ is any unramified extension of $\mathbb{Q}_5$ and $E/K$ is totally ramified with $e = [E : K]$ coprime to $p$, then $E \cong K((u \cdot 5)^{1/e})$ for some unit $u \in \mathcal{O}_K^\times$. If we allow wild ramification ($p \mid e$), then there are many more possible extensions, but the possibilities for tamely ramified extensions of local fields are very restricted.