This question could have been asked by several users. Here is my attempt. This could possibly be completely in the wrong direction. So I attempt to show $F(\{x,y,z\}) $ is a subgroup of $F(\{a,b\})$. My intention is to exploit the universal mapping property (UMP) (and it could be in the wrong direction but anyway I'll give it a try for reasons stated below). To begin with, we have the natural injection $i:\{x,y,z\}\to F(\{x,y,z\})$ s.t $i(x)=x,i(y)=y,i(z)=z$ and the map $f:\{x,y,z\}\to F(\{a,b\})$ given by $f(x)=a^2,f(y)=ab,f(z)=b^2$. So by UMP of $F(\{x,y,z\})$ there is a unique $\phi: F(\{x,y,z\})\to F(\{a,b\})$ (homomorphism) such that $f=\phi \circ i$ i.e $\phi(x)=a^2,\, \phi(y)=ab,\, \phi(z)=b^2$. So $F(\{x,y,z\})/ker(\phi)$ is (isomorphic to) a subgroup of $F(\{a,b\})$ and hence free.
So can we conclude from this that $ker(\phi)$ is trivial?? We will be done if this happens.
A more general question (possibly wrong) can be formulated as follows :
If $\psi : F(X)\to F(Y)$ is a surjective homomorphism between free groups, is $ker(\psi)$ trivial??
(Edit): So may be a better way to formulate the general question would be : If $F(X)$ is a free group and a factor group $F(X)/N$ is also free then does that imply the normal subgroup $N$ to be trivial?? I am having a feeling that the answer to this question is in affirmative but I am nowhere near a solution.
P.S: There could be some possible directions to prove this using fundamental groups, but my algebraic topology is quite rusty (hence didn't tag it) so I would prefer algebraic attempts. But anyways, all ideas are welcome and thanks in advance. Also I am not really good at finding counterexamples in questions related to free group, so if this question is stupid, please ignore.
I'll make atry to prove that $ker(\phi)=1$. Let $T=\langle a^2,ab,b^2 \rangle$. It suffies to show that $a^2,ab,b^2$ freely generate $H$. Equivalently the elements $$x=a^2(ab)^{-1}=a^2b^{-1}a^{-1},\ y=ab,\ z=b^2(ab)^{-1}=ba^{-1}$$ freely generate $H$. In any non-cancelling product of two of $x^{\pm1},y^{\pm1},z^{\pm1}$ there is always some $a^{\pm1}$ or $b^{\pm1}$. Hence no such word reduces to the identity and it follows that $H$ is freely generated by $x,y,z$.
For your second question, if $\psi: F(X)\to F(Y)$ is surjective and $ker(\psi)$ is trivial then it is an isomorphism. But I can't find a surjective homomorphism $F(s,t,w)\to F(k,l)$ which is not injective (since a group on $3$ letters is not isomorphic to a group on $2$ letters).