 Solution

Suppose $x,y$ have order three but $x\ne y,y^2$. Their cyclic subgroups have trivial intersection: any common nonidentity element would generate both subgroups and force $x=y$ or $y^2$. The nine products $x^i{y}^j$, $0\le i,j<3$, are distinct. Indeed $x^i{y}^j=x^k{y}^{\ell }$ implies $x^{i-k}=y^{\ell -j}$ in that intersection, so $i=k$ and $j=\ell$ modulo three. Nine distinct elements cannot fit in a group of order six, proving the claim without assuming commutativity.