f'(x,y,z) = [3·x^2 + 6·y, 6·x + 2·y, 2·z + 2] = [0, 0, 0]
f''(x, y, z) = [6·x, 6, 0; 6, 2, 0; 0, 0, 2]
2·z + 2 = 0 --> z = -1
6·x + 2·y = 0 --> y = -3·x
3·x^2 + 6·(-3·x) = 0 --> x = 6 ∨ x = 0
Lösung sind daher
(x = 6 ∧ y = -18 ∧ z = -1) ∨ (x = 0 ∧ y = 0 ∧ z = -1)
f''(6, -18, -1) = [36, 6, 0; 6, 2, 0; 0, 0, 2] --> positiv definit --> Minimum
f''(0, 0, -1) = [0, 6, 0; 6, 2, 0; 0, 0, 2] --> indefinit --> Sattelpunkt