You can always rotate the circle so that ray $z$ lies on top of where you would consider $0^\circ$ to normally be. This rotation wouldn't affect the angles. Then you just get the angle between $x$ and $y$ by subtracting the difference as usual.
A simple proof of this is as follows. Let $x_z$ be the angle of $x$ in terms of $z$ and let $y_z$ be the angle of $y$ in terms of $z$. Let's also use $x,y,z$ to denote the angles of the respective rays according where you consider $0^\circ$ to be. We want to find $x-y$.
$$x-y=(x-z)-(y-z)=x_z-y_z$$
You can always rotate the circle so that ray $z$ lies on top of where you would consider $0^\circ$ to normally be. This rotation wouldn't affect the angles. Then you just get the angle between $x$ and $y$ by subtracting the difference as usual.
A simple proof of this is as follows. Let $x_z$ be the angle of $x$ in terms of $z$ and let $y_z$ be the angle of $y$ in terms of $z$. Let's also use $x,y,z$ to denote the angles of the respective rays according where you consider $0^\circ$ to be. We want to find $x-y$. $$x-y=(x-z)-(y-z)=x_z-y_z$$