It's a bit late answer, but there is a nice prove :).
Denote $u = x^3 = y^5$, so $u$ is in center of $G$. Add new symbol $q$ and state that it commutes with other symbols and $q^{-30}=u$, so we obtain new group $E$ isomorphic to $G \times C_{30}$. Now denote $a = q^{10}x, b=q^{6}y, c=q^{15}yx$. It is easy to check that $a^3=b^5=c^2=1$ in $E$. Denote $Q_1 = q^{-1}ba, Q_2 =q^{-1}ab$. You can check that $Q_1^2 = Q_2^2 = 1$. The next holds: $Q_1Q_2 = q^{-6}(b^2a^2)b(b^2a^2)^{-1}$. And hence $(Q_1Q_2)^5=q^{-30}=u$. But $(Q_2Q_1)^5 = Q_1(Q_1Q_2)^5Q_1 = u$ so $u^2=(Q_1Q_2)^5(Q_2Q_1)^5=1$ in $E$. Clearly, $u^2=1$ should holds in $G$ too.
The idea of this proof is from this article: http://link.springer.com/article/10.1007%2Fs11253-012-0689-yoof