From the given conditions, we have $$x+y=6\quad\text{and}\quad xy=z^2+9$$ Suppose $x\ne y$, then they must be the two real roots of $$u^2-6u+(z^2+9)=0$$ It follows that this equation's determinant must be non-negative: $$\Delta = 36-4(z^2+9)\ge 0 \implies z^2 \le 0 \implies z=0 \implies \Delta=0$$ This implies $x=y$. When $x=y$, we find $(x,y,z)=(3,3,0)$ satisfies the conditions. Therefore we conclude $x=y$ must hold.