f(x) = -x^3 - k·x^2 + (k - 1)·x
f'(x) = -3·x^2 - 2·k·x + k - 1 = 0 --> k = (3·x^2 + 1)/(1 - 2·x)
y = -x^3 - (3·x^2 + 1)/(1 - 2·x)·x^2 + ((3·x^2 + 1)/(1 - 2·x) - 1)·x
y = (x^4 - 2·x^3 - 1·x^2)/(2·x - 1)
f''(x) = - 6·x - 2·k = 0 --> k = -3·x
y = -x^3 - (-3·x)·x^2 + (-3·x - 1)·x
y = 2·x^3 - 3·x^2 - x