(answer adapted to some of the comments)
The order of $H$ is the order of its generator $x^6,$ i.e., 4. Therefore the order of $G/H$ is 6: there are 6 different cosets of the form $x^iH.$
We can choose the first powers of $x$ as representatives of these cosets:
The order of an individual coset $x^iH$ is the smallest nonzero natural number $n$ such that $i.n$ is a multiple of 6.
Now at least one of the elements of $G/H$ turns out to have its order equal to the order of $G/H$ itself, implying that $G/H$ is cyclic.
A cyclic group is generated by an individual element $g$ iff the order of that element is the order of the group.