Let the coordinates of the point in the first quadrant (note there are two points 3 units away, the other in the third quadrant) be $(x,y)$.
Since the slope of the ray is $1\over2$, and since slope is "rise/run": $ \tag{1}{1\over2}={y\over x}. $ By the Pythagorean Theorem $ \tag{2}x^2+y^2=9. $
We need to solve the system of equations (1) and (2).
Solving (1) for $y$ gives $ \tag{3}y={x\over 2}. $ Replacing $y$ in (2) with ${x\over 2}$ gives $ x^2+\bigl({\textstyle{x\over2}}\bigr)^2=9; $ or $ {5x^2\over 4}=9. $ Solving the above for positive $x$ gives $x^2={9\cdot4 \over 5}$; whence $x=6/\sqrt5$. And then from (3), $y=3/\sqrt5$ (the point in the third quadrant is $x=-6/\sqrt5$, $y=-3/\sqrt5$).