f(x) = x2/3(2 - x)
I would first distribute to make it easier to derive (otherwise we would have to use the product rule and that can get messy).
f(x) = 2x2/3 - x5/3 (remember x is really x1 or x3/3 in this case, so x2/3 times x3/3 is x5/3)
Now use the power rule:
f'(x) = 2(2/3)x-1/3 - (5/3)x2/3 = (4/3)x-1/3 - (5/3)x2/3
Now do the same thing to f'(x) to get f"(x):
f"(x) = (4/3)(-1/3)x-4/3 - (5/3)(2/3)x-1/3 = (-4/9)x-4/3 - (10/9)x-1/3