g'(x) = 6x^2 + 6x - 36 = 0
= x^2 + x - 6 = 0
= (x + 3)(x - 2) = 0
so you have critical points at x = -3 and x = 2
since x = -3 we can ignore this one. We take the second derivative to get
g''(x) = 2x + 1
g''(2) = 5 > 0. Since this is a minimum, we check the endpoints x = 0 and x = 3 and plug them into g(x) to get
g(0) = 2
g(3) = 2 (3)^3 + 3 (3)^2 - 36(3) + 2
= 2(27) + 27 - 108 + 2
= 54 + 27 - 108 + 2
= 81 - 108 + 2 = -25
Therefore the absolute maximum of the function in the interval [0,3] is at x = 0 where g(0) = 2.