If $G$ has exponent 3, then $G$ is a $2$-Engel group. I think this was proven by Kappe and Kappe.
I want to show that if $G$ has exponent 3, this is, for any $g\in G$ we have that $g^3=1$. Then $G$ is a 2-Engel group, in other words, $[x,y]=x^{-1}y^{-1}xy$ commutes with $y$, or  equivalently  $x$ commutes with $x^y=y^{-1}xy$.
My idea is the following, we know that $(xy^{-1})^3=1$, so $xy^{-1}xy^{-1}=yx^{-1}$.
From this equality, I would like to obtain the equality $xx^y=x^yx$, which would imply that $G$ is a $2$-Engel group. But I have tried obtaining this result with no success.
Any help would be appreciated.
 A: This was proven independently by a number of people. It is often attributed to F.W. Levi and van der Waerden, in their paper Über eine besondere Klasse von Gruppen, Abh. Math. Sem. Hamburg 9 (1933), 154-158; I think it is also proven in  Levi's later paper, Groups in which the commutator operation satisfies certain algebraic conditions, J. Indian Math. Soc. 6 (1942), 87-97.
From $(xy)^3 = e$, we obtain $yxy=x^{-1}y^{-1}x^{-1}$ for any $x$ and $y$. Hence
$$\begin{align*}
a^{-1}baba^{-1} &=
(a^{-1}ba^{-1}) aaa (a^{-1}ba^{-1})\\
&= (a^{-1}ba^{-1})(a^{-1}ba^{-1})\\
&= (b^{-1}ab^{-1})(b^{-1}ab^{-1})\\
&= b^{-1}(ab^{-2}a)b^{-1}\\
&= b^{-1}(b^2a^{-1}b^2)b^{-1}\\
&= ba^{-1}b.
\end{align*}$$
Thus, $a^{-1}baba^{-1} = ba^{-1}b$, and therefore
$$(a^{-1}ba)b = b(a^{-1}ba).$$
Thus, every element $b$ commutes with any conjugate $a^{-1}ba$ of $b$.
That means that $y$ commutes with $[y,x] = y^{-1}(x^{-1}yx)$, since it commutes with both $y^{-1}$ and with $x^{-1}yx$. Thus, $[y,x,y]=[[y,x],y]=e$. And it also commutes with $[y,x]^{-1}=[x,y]$, so $[x,y,y]=e$ as well. Hence, a group of exponent $3$ is $2$-Engel.
