$$ f (x,y,z) = xy - z^4 -2(x^2 + y^2 - z^2) $$
$$NB :0=x^2 + y^2 + 2z^2 - 8 $$
$$ \Lambda (x,y,z,\lambda) = xy - z^4 -2(x^2 + y^2 - z^2) - \lambda(x^2 + y^2 + 2z^2 - 8) $$
$$ \Lambda (x,y,z,\lambda) = xy - z^4 -2x^2 -2y^2 +2 z^2 - \lambda(x^2 + y^2 + 2z^2 - 8) $$
und jetzt die partiellen Ableitungen nach x,y,z und Lambada bilden:
$$ \Lambda (x,y,z,\lambda) = xy - z^4 -2x^2 -2y^2 +2 z^2 - \lambda(x^2 + y^2 + 2z^2 - 8) $$
---
$$ \frac {\partial \Lambda (x,y,z,\lambda)}{\partial x} = y -4x - 2x \lambda$$
$$ \frac {\partial \Lambda (x,y,z,\lambda)}{\partial y} = x -4y - 2y \lambda$$
EDIT:\\
[$$ \frac {\partial \Lambda (x,y,z,\lambda)}{\partial z} = - 4z^3 +4 z - 2 z \lambda$$]
korrigiert:$$ \frac {\partial \Lambda (x,y,z,\lambda)}{\partial z} = - 4z^3 +4 z - 4 z \lambda$$
$$ \frac {\partial \Lambda (x,y,z,\lambda)}{\partial \lambda} = x^2 + y^2 + 2z^2 - 8 $$
Dann diese Nullsetzen und versuchen zu lösen ...
$$0 = y -4x - 2x \lambda$$
$$ 0 = x -4y - 2y \lambda$$
$$0 = - 4z^3 +4 z - 2 z \lambda$$
$$ 0 = x^2 + y^2 + 2z^2 - 8 $$