a)
x * y = 1 --> y = 1/x
f(x, y) = x^3 - x·y + y^3
f(x) = x^3 - x·(1/x) + (1/x)^3 = x^3 + 1/x^3 - 1
f'(x) = 3·x^2 - 3/x^4 = 0
x1 = -1 und y1 = 1/(-1) = -1
x2 = 1 und y2 = 1
b)
f(x, y) = 4·x^2 + 3·x·y + 2·y^2 + 7·x - 6·y - 6
fx'(x, y) = 8·x + 3·y + 7 = 0
fy'(x, y) = 3·x + 4·y - 6 = 0
Wir lösen das Gleichungssystem und erhalten [x = -2 ∧ y = 3]