Any Subgroup containing commutator subgroup is normal.

If $g\in G$ and $h\in H$, then $ghg^{-1}h^{-1}=h'$, for some $h'\in H$ (since $H$ contains the commutator subgroup). But then $ghg^{-1}=h'h\in H$. Therefore, $gHg^{-1}\subset H$.

$G'$ is certainly normal in $G$, and $G/G'$ is Abelian. Every subgroup of an Abelian group is normal. But $H/G'$ is a subgroup of $G/G'$ so $H/G'$ is normal in $G/G'$. Therefore, by the third isomorphism theorem for groups, $H$ is normal in $G$.