令 $F=z^3-2xz+y$，则 $F_x=-2z$，$F_y=1$，$F_z=3z^2-2x$。$\dfrac{\partial z}{\partial x}=-\dfrac{F_x}{F_z}=\dfrac{2z}{3z^2-2x}$，$\dfrac{\partial z}{\partial y}=-\dfrac{1}{3z^2-2x}$。

$\dfrac{\partial^2 z}{\partial x\partial y}=\dfrac{\partial}{\partial y}\left(\dfrac{2z}{3z^2-2x}\right)=\dfrac{2\dfrac{\partial z}{\partial y}(3z^2-2x)-2z\cdot 6z\dfrac{\partial z}{\partial y}}{(3z^2-2x)^2}=\dfrac{2\dfrac{\partial z}{\partial y}(-3z^2-2x)}{(3z^2-2x)^2}$。

将 $\dfrac{\partial z}{\partial y}=-\dfrac{1}{3z^2-2x}$ 代入，得 $\dfrac{6z^2+4x}{(3z^2-2x)^3}$。