Given SQR = 60^ and SPR = y.

The angles $\angle SQR$ and $\angle SPR$ are both subtended by the arc SR at the circumference. Angles subtended by the same arc at the circumference are equal.
Given $\angle SQR = 60^\circ$ and $\angle SPR = y$.

$$y = \angle SQR$$ $$y = 60^\circ$$

The angle $\angle SOR$ is the angle subtended by arc SR at the center O. The angle $\angle SQR$ is the angle subtended by arc SR at the circumference. The angle at the center is twice the angle at the circumference subtended by the same arc.
Given $\angle SOR = 3x$ and $\angle SQR = 60^\circ$.

$$\angle SOR = 2 \times \angle SQR$$ $$3x = 2 \times 60^\circ$$ $$3x = 120^\circ$$

Divide both sides by 3.

$$x = \frac{120^\circ}{3}$$ $$x = 40^\circ$$

Now substitute the values of $x$ and $y$ into the expression $x+y$.

$$x+y = 40^\circ + 60^\circ$$ $$x+y = 100^\circ$$

The value of $x+y$ is $\boxed{\text{}100^\circ\text{}}$.