National Math Olympiad

The second equation implies $2xy=3yz+xz$. Multiplying the first equation respectively by $z$, $x$ and $y$, we get $z^2=3xz+2zy$, $3x^2=zx-2xy$ and $2y^2=zy-3xy$. So, $$5x^2-4y^2-z^2=\frac{5}{3}(zx-2xy)-2(zy-3xy)-(3xz+2zy)=\frac{4}{3}(2xy-3zy-xz)=0.$$