My question is exactly the one written in the title.
My first idea was to find the subgroup lattice of $\mathbb{Z}_2\times \mathbb{Z}_2\times\mathbb{Z}_3\times \mathbb{Z}_3$, and separate the once containing the diagonal $\{(a, a) \mid a\in\mathbb{Z}_6\}\simeq\mathbb{Z}_6.$ But, it feels like there should be a more systematic way of doing this.
Let's call the group $G$, and the diagonal $D$. Then $D$ is a normal subgroup of $G$, and so there is a bijective correspondence between subgroups of $G$ containing $D$, and the subgroups of the quotient $G/D$. Note that $G/D$ is an abelian group of order $6$, and so $G/D\cong\mathbb{Z_6}$. So it follows that there exactly $4$ such subgroups.
And now note that for each divisor $d$ of $6$ there is such a subgroup:
$H_d=\{(a,b)\in G: \text{$a-b$ is divisible by $d$}\}$
Check that $H_d$ is well defined, and is a subgroup containing $D$. So these are the $4$ required subgroups.