In the definition of a presentation of a group by generators and relations, we take the quotient of the free group on the generators by the smallest normal subgroup generated by the relations. This has 2 consequences: (a) normality of the subgroup we quotient by is not a problem; (b) the subgroup we quotient by is huge.
This subgroup contains not just $xyx^{-1}y^{-1}$ and its powers, but (because of normality) it also contains anything of the form $wxyx^{-1}y^{-1}w^{-1}$ where $w$ is an arbitrary word in the free group on $x,y$, as well as any product of any finite number of words of that form (and their inverses), and conjugates of those products by any words, and any products of those things, and their conjugates, and so on.
So, the free group is huge, but so is the group we quotient by, so it's reasonable to expect that we get something "small" like $\mathbb{Z} \times \mathbb{Z}$.
In fact, the best way to think about groups presented by generators and relations is not to think explicitly about what we quotient by. Intuitively, the reason this quotient group is $\mathbb{Z} \times \mathbb{Z}$ is because the relation $xyx^{-1}y^{-1}=1$ implies $xy=yx$, which allows us to rearrange the letters in a word inside the free group into the form $x^a y^b$. The product becomes $(x^a y^b) (x^c y^d) = x^{a+c} y^{b+d}$ so the group is $\mathbb{Z} \times \mathbb{Z}$.
A precise argument use the following universal property of quotient groups: Let $H$ is a normal subgroup of $G$. Suppose $K$ is an arbitrary group for which we have a homomorphism $f : G \to K$ with kernel containing $H$. Then there exists a unique homomorphism $\overline{f}: G/H \to K$ such that the following composition is $f$: $$G \to G/H \to K.$$ Here the first map in the composition is the natural projection map from $G$ to $G/H$, and the second is $\overline{f}$.
Intuitively, this universal property is saying that $G/H$ is the "largest" group for which we have a surjective homomorphism from $G$ with kernel containing $H$, because any other group with this property "factors through" $G/H$, as in the 2-step composition above.
All of this means that in your example problem, we must show that $\mathbb{Z} \times \mathbb{Z}$ is the largest group which is generated by 2 elements $x', y'$ with the property $x' y' x'^{-1} y'^{-1} = 1$. To prove that this largest group is $\mathbb{Z} \times \mathbb{Z}$ means that:
(1) there is a surjection $F_2 \to \mathbb{Z} \times \mathbb{Z}$.
(2) for any other group $G$ has this property (with 2 elements $x'', y''$ such that $x'' y'' x''^{-1} y''^{-1} = 1$), there is a map $\overline{f}$ from $\mathbb{Z} \times \mathbb{Z}$ to $G$ such that this composition takes takes $x$ to $x''$ and $y$ to $y''$: $$F_2 \to \mathbb{Z} \times \mathbb{Z} \to G$$
(where the first map is the surjection in (1) and the second is $\overline{f}$.)
This is much easier because we don't have to think directly about all the words in the complicated normal subgroup generated by $xyx^{-1}y^{-1}$. To prove (1), we find 2 elements $x', y'$ that generate $\mathbb{Z} \times \mathbb{Z}$ and satisfy $x' y' x'^{-1} y'^{-1} = 1$. Then by the universal property of the free group, there is a unique map $F_2 \to \mathbb{Z} \times \mathbb{Z}$ taking $x \mapsto x'$ and $y \mapsto y'$. So we get our map (1). Then for (2), we have an arbitrary group $G$ with 2 elements $x'', y''$ satisfying our relation. By using that relation, we can reduce any element of $G$ to the form $(x'')^a (y'')^b$. Check that we must send $(a,b) \mapsto (x'')^a (y'')^b$ to get a homomorphism as in (2).
This was long and I've left out many details at the end, but I hope it was helpful.
Best Answer
The commutator of any two pairs $(a,b),(c,d)$ is $(0,[b,d])$. So the commutator subgroup is $\{0\}\times [S_3, S_3]=\{0\}\times A_3$ where $A_3$ consists of the identity and two $3$-cycles and is the cyclic subgroup of $S_3$ of order $3$.