Найдите точки экстремума функции: у=х^3/3-х^2-3 х

Y' (x) = (3x² * (3-x²-3x) - x³ * (-2x-3)) / (3-x²-3x) ² = (9x²-3x^4-9x³+2x^4+3x³) / (3-x²-3x) ²=

(-x^4-6x³+9x²) / (3-x²-3x) ²=x² (9-6x-x²) / (3-x²-3x) ²=0

x² (9-6x-x²) = 0

x=0

9-6x-x²=0

D=36+36=72 √D=6√2

x1 = (6-6√2) / - 2=3√2-3

x2 = (6+6√2) / - 2=-3√2-3