= Solution
For roots $r_1,r_2,r_3$, the discriminant is
$$
\Delta(f)=\prod_{i<j}(r_i-r_j)^2,
$$
and for $f=X^3+pX+q$,
$$
\boxed{\Delta=-4p^3-27q^2}.
$$
If $f$ is irreducible, its Galois group is transitive in $S_3$: it is $A_3\cong C_3$ when $\Delta$ is a square in $K$, and $S_3$ otherwise. If $f$ has one root in $K$ and an irreducible quadratic factor, the group is $C_2$; if it splits, it is trivial. A repeated-root depressed cubic already has all roots in $K$, so adds no case.
For $X^3+3X+1$, the rational-root test proves irreducibility and
$$
\Delta=-135=-3^3\cdot5
$$
is not a rational square. Thus $G\cong S_3$. By the <Galois correspondence>[fundamental theorem of Galois theory], the complete subfield list is
$$
\mathbb Q,quad L,quad
\mathbb Q(r_1),\mathbb Q(r_2),\mathbb Q(r_3),quad
\mathbb Q(\sqrt{-135})=\mathbb Q(\sqrt{-15}),
$$
where the three cubic fields fix the three order-two subgroups and the quadratic field fixes $A_3$.
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