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.)