So, let's first find the slope of the second question:
[tex] 2y=x+3 \implies\\
y=\frac{x}{2}+\frac{3}{2} [/tex]
Our second equation has a slope of one half, so we know that a line perpendicular to it will have the negative reciprocal slope of that, so -2. We have this now as our equation:
[tex] y=-2x+b [/tex]
We are given the point (3,-6), so let's plug the x and y in and find our b:
[tex] -6=-2(3)+b \implies\\
-6=-6+b \implies\\
0=b [/tex]
Since our b is 0, our equation is simply:
[tex] y=-2x [/tex]
