I have shown that $f(t)$ is irreducible, though I could not find an easier way to do this than by showing factorization into one polynomial of degree $4$ and one of degree $2$, or into two of degree $3$ were both impossible - this seems clunky, so I am surprised if it is necessary.
I've considered the splitting field of $g(t)=t^2-2t-1$, which is an intermediate extension $M$ such that $Gal(M:K)=C_2$. Since $M$ and $L$ are both normal over $K$, we have $Gal(M:K)=Gal(L:K)/M^*$, where $M^*$ is the subgroup of all $M$-automorphisms; I expect this to be useful. I have a few ideas from here, e.g. if $\beta_1,\beta_2$ are the roots of $g(t)$ I believe $L$ is the splitting field of $(x^3-\beta_1)(x^3-\beta_2)$ over $M$, but I am not managing to link these up into an actual description of the Galois group. Is this the right direction?
Thank you in advance for reading through this!
I'll highlight some main steps, and leave some of the details and justifications for you to verify.
$(1)$ Let $f(x):=x^6-2x^3-1$ and $g(t):=t^2-2t-1$, so $x^3=t$. Then the roots of $g(t)=t^2-2t-1$ are $1\pm\sqrt 2$ so that the roots of $f(x)$ are $\sqrt[3]{1\pm\sqrt 2},~\omega\sqrt[3]{1\pm\sqrt 2},~\omega^2\sqrt[3]{1\pm\sqrt 2}$ with $\omega:=e^{2\pi i/3}$.
$(2)$ So the splitting field of $f(x)$ over $\Bbb Q$ is $M:=\Bbb Q(\sqrt[3]{1+\sqrt 2},\omega)$ which is a degree $12$ extension over $\Bbb Q$ since $[\Bbb Q(\sqrt[3]{1+\sqrt 2}):\Bbb Q]=6$ by irreducibility of $f(x)$ and $[\Bbb Q(\omega):\Bbb Q]=2$. Note that we omit $\sqrt[3]{1-\sqrt 2}$ from the generators of $M$ since $\sqrt[3]{1+\sqrt 2}=-(\sqrt[3]{1-\sqrt 2})^{-1}$. For complete rigour, you will need to verify this degree computation and the claim in the previous sentence.
$(3)$ It follows that $|\operatorname{Gal}(f/\Bbb Q)|=12$ so the Galois group of $f(x)$ over $\Bbb Q$ is $D_6$ (dihedral of order $12$) or $C_6\times C_2$ up to isomorphism, as the Galois group is a transitive subgroup of $S_6$.
$(4)$ We can show that the Galois group is not abelian, and thus $\operatorname{Gal}(f/\Bbb Q)\cong D_6$. To show non-abelian, use the fundamental theorem and exhibit a non-normal subfield of $M$.