There is a homomorphism $\langle x,y\rangle\to\langle x\rangle$ of free groups, sending $y$ to $1$. We can combine this with the other obvious homomorphism to get a surjective homomorphism $$ \langle x,y\rangle \to \langle x\rangle\times\langle y\rangle $$ The kernel is the commutator subgroup, and it is freely generated by the elements $[x^i,y^j]$ for $i,j\in\mathbb{Z}$ with $i,j\neq 0$.

We can define a homomorphism
$$ \langle x,y,z\rangle \to \langle x,y\rangle \times \langle x,z\rangle \times \langle y,z\rangle $$
in a similar way, and let $K$ denote the kernel. In some notes of Vershinin I have seen this called the *fat commutator subgroup*. It is a subgroup of a free group and so must be free.

Question: Is there a known basis?

Let $X$ denote the smallest normal subgroup of $\langle x,y,z\rangle$ containing $x$, or equivalently the kernel of the map $\langle x,y,z\rangle \to \langle y,z\rangle$. Define $Y$ and $Z$ similarly, so $K=X\cap Y\cap Z$. Put $$ A = \{[u,[v,w]] \;:\; (u,v,w)\in X\times Y\times Z\}. $$ It is easy to see that $A\subseteq K$. There is a comment of Vershinin which might mean that $A$ generates $K$, or it might mean something a bit more complicated. In any case, the document that I found did not contain a proof or reference. Also, I think that $A$ is too big to generate $K$ freely.