To find the max speed you need to find the time derivative:

v(t) = 6 cos(3t)i- 3 sin(3t)j(This is the velocity, incidentally. You gave the displacement function, a vector.)

What is the speed of this particle?

v(t) = sqrt( v_x ^2 + v_y ^2) = sqrt(36 cos^2(3t) + 9 sin^2(3t))

v(t) = sqrt(27 cos^2(3t) + 9)

So maximum speed will occur when a(t) is 0:

a(t) = (-162 sin(3t) cos(3t)) / (2 [sqrt(36 cos^2(3t) + 9 sin^2(3t))]) = 0

Or when -162 sin(3t) cos(3t) = 0.

This equation has a quite simple solution: t = 0 can be seen by inspection. I leave it to you to show that this is a maximum speed, not a minimum.

v(0) = sqrt(27 + 9) = sqrt(36) = 6 (in whatever units you are using.)

-Dan