Wait, differentiate or integrate?
I'll assume integration.
cos^(4)3t sin3t dt
You got the first step right. take u = 3t -> du = 3dt -> du/3 = dt. You can pull the 1/3 constant out of the integral after substitution
(1/3) * the integral cos^(4)u sinu du.
Now, use substitution again for this new integral. In this case, the substitution is simple, but in other trig forms like the one above, you may have to use some identities before substitution.
let v = cos u
dv = -sin u du
-dv = sin u du
You can pull the negative sign out of the integral after substitution as you had done with 1/3
-(1/3) * the integral of v^4 dv
-(1/3) * v^5/5
-(1/3) * cos^(5) u /5 (substitute back into v)
-(1/3) * cos^(5) 3t /5 (substitute back into u)
Hope that helps.