CEMC Pascal

Since $\angle X M Z=30^{\circ}$, then $\angle X M Y=180^{\circ}-\angle X M Z=180^{\circ}-30^{\circ}=150^{\circ}$.
Since the angles in $\triangle X M Y$ add to $180^{\circ}$, then
$$\angle Y X M=180^{\circ}-\angle X Y Z-\angle X M Y=180^{\circ}-15^{\circ}-150^{\circ}=15^{\circ}$$
(Alternatively, since $\angle X M Z$ is an exterior angle of $\triangle X M Y$, then $\angle X M Z=\angle Y X M+\angle X Y M$ which also gives $\angle Y X M=15^{\circ}$.)
Since $\angle X Y M=\angle Y X M$, then $\triangle X M Y$ is isosceles with $M X=M Y$.
But $M$ is the midpoint of $Y Z$, and so $M Y=M Z$.
Since $M X=M Y$ and $M Y=M Z$, then $M X=M Z$.
This means that $\triangle X M Z$ is isosceles with $\angle X Z M=\angle Z X M$.
Therefore, $\angle X Z Y=\angle X Z M=\frac{1}{2}\left(180^{\circ}-\angle X M Z\right)=\frac{1}{2}\left(180^{\circ}-30^{\circ}\right)=75^{\circ}$.