In my opinion the easiest way to do this is to expand the function as
t2/3(2t-1) = 2t5/3 - t2/3
A graph will show that there are only 2 critical points and they should be able to be found by setting the derivative = 0.
Nope, I am wrong.
Compute the derivative by the product rule.
f'(t) = 2t2/3 + (2t-1)(2/3)t-1/3 = 0
This has 2 solutions: t=0 and (if you multiply through by t1/3, you will easily see that the other critical point is at t =1/5.