Critical points on f are defined as anywhere f'(x) = 0 or is undefined. Since f a polynomial, f' is defined everywhere, so we can simply differentiate f (using power rule) to get f', set it = 0, and solve:
f'(x) = 8x3 - 24x2 = 0
8x2(x - 3) = 0
x = 0 or x = 3 are the only two critical points for f.
Thinking about a sign chart for f' will allow us to determine which type these are. f'(x) is an odd degree poly with a positive leading coefficient, so it starts in Q III and finishes in QI. Thus, f' is negative for x < 0. At x = 0, it has a root of even multiplicity, so its graph has a tangent root there. So f' returns to being negative for 0 < x < 3, before crossing the x-axis at x = 3 and remaining positive for 3 < x.
f(x) has a flat point of inflection at (0,1) and a local minimum at (3,-53).
You can and should use Desmos or another graphing utility to confirm this analysis.