Calculus: Early Transcendentals, Chapter 4, 4.1, Section 4.1, Problem 43

You need to evaluate the critical numbers of the function, hence, you need to evaluate the soutions to the first derivative, such that:
f'(x) = 0
You need to find the first derivative using the product rule:
f'(x) = (x^2)'*(e^(-3x)) + x^2*(e^(-3x))'
f'(x) = 2x*(e^(-3x)) + x^2*(e^(-3x))*(-3x)'
f'(x) = 2x*(e^(-3x)) - 3x^2*(e^(-3x))
You need to factor out (e^(-3x)):
f'(x) = (e^(-3x))*(2x - 3x^2)
You need to solve for x the equation f'(x) = 0:
(e^(-3x))*(2x - 3x^2) = 0
Since e^(-3x) >0, then 2x - 3x^2 = 0.
You need to factor out x, such that:
x(2 - 3x) = 0 => x = 0
2 - 3x = 0 => 2 = 3x => x = 2/3
Hence, evaluating the critical values of the function yields x = 0, x = 2/3.