a question about fixed-point-free automorphism

Let G be a finite group with a fixed-point-free automorphism a of order 3. Prove that [x,y,y]=1 for all x,y in G.

I will summarize the proof from the book "Theory of Groups of Finite Order" by W. Burnside. It would not be very easy to think of this if you had not seen it before!

With your notation, conjugation by $a$ is inducing a fixed-point-free automorphism of order $3$ of $G$. You know that $x$ commutes with $a^{-1}xa$ and similarly with $axa^{-1}$.

If you conjugate $(axa^{-1})x(a^{-1}xa)$ by $a$ then you get $x(a^{-1}xa)(axa^{-1})$, which is the same element, so $(axa^{-1})x(a^{-1}xa) = 1$, and hence $x \in \langle a^{-1}xa,axa^{-1} \rangle$.

Since $a$ acts fixed-point-freely, the $|G|$ elements $g^{-1}ag$ for $g \in G$ are all distinct, and they are all of the form $g'a$ for some $g' \in G$, so we have $\{ g^{-1} a g : g \in G \} = \{ga : g \in G \}$.

You are trying to prove that $[[x,y],y]=1$ for all $x,y \in G$, which is equivalent to proving that $y$ commutes with $x^{-1}yx$.

By the above, we have $xa = g^{-1}ag$ for some $g \in G$. Now $g^{-1}ag$ also has order $3$ and acts fixed-point-freely on $G$, so $y$ commutes with $(g^{-1}a^{-1}g)y(g^{-1}ag) = a^{-1}x^{-1}yxa$. Similarly, $y$ commutes with $ax^{-1}yxa^{-1}$. But we saw above that $x^{-1}yx \in \langle a^{-1}x^{-1}yxa,ax^{-1}yxa^{-1} \rangle$, so $y$ commutes with $ x^{-1}yx$, and we are done.