Well, let's see ...
You said that f(t) = 3t³ - 3t² + t - 1
I have to assume that f(t) is the position of the praticle at time t .
Then the object's speed is the first derivative. That's s(t) = 9t² - 6t + 1
Speed is zero when S(t) = 9t² - 6t + 1 = 0
Solve that ugly thing with the quadratic formula, and you'll get t = 1/3 .
The object's acceleration is the second derivative. That's A(t) = 18t - 6 .
Acceleration is zero when A(t) = 18t - 6 = 0 .
Solve this thing with elementary algebra, and you'll get t = 1/3 .
Ordinarily we'd be surprised to encounter a praticle whose speed and acceleration are both apparently zero at the same time. This would cause us discomfort, and we would check our work over and over about six times, but we'd keep coming up with the same answers. We'd slink away then, more uncomfortable than ever and with the dark suspicion that this praticle must have some kind of goofy motion.
THAT happens to be exactly what's going on. The praticle's position f(t) is a third-order polynomial. If you look at the graph, you find that its first and second derivatives are both zero at t=1/3 . So everything is sunny and wonderful . . . not only fine, but dandy as well !