$\mathbb{Q}({\zeta }_3,\sqrt[3]{2},\sqrt[3]{3})$ is a Kummer extension with quotient $(\mathbb{Z}\mathrm{/}3{)}^2$

The field $F$ contains ${\mu }_3$, so adjoining cube roots is exactly the Kummer situation of [L1]. To show that the classes of $2$ and $3$ are independent modulo cubes, suppose $2^a{3}^b=c^3\text{in }{F}^{\times }\mathrm{.}$ Taking norms from $F$ to $\mathbb{Q}$ gives $2^{2a}{3}^{2b}=N_{F\mathrm{/}\mathbb{Q}}(c{)}^3\mathrm{.}$ The left side is a rational cube only when $3\mid 2a$ and $3\mid 2b$, hence only when $3\mid a$ and $3\mid b$. Therefore the classes of $2$ and $3$ each have order $3$ and generate a subgroup isomorphic to $(\mathbb{Z}\mathrm{/}3{)}^2$.

By [L1], the field generated by the corresponding cube roots is a Kummer extension with Galois group $\mathrm{Hom}((\mathbb{Z}\mathrm{/}3{)}^2,{\mu }_3)\cong (\mathbb{Z}\mathrm{/}3{)}^2\mathrm{.}$ Since $K$ is exactly that field, the stated conclusion follows.