Apply implicit function theorem on $z^3 - 3xyz = 1$ and $z = z(x,y)$

What is $z'$? We assume that $z$ is a function of $x$ and $y$, so it has partial derivatives $\frac{\partial z}{\partial x}$ and $\frac{\partial z}{\partial y}$. They can be calculated by the implicit function theorem using the partial derivatives of $F$, which are equal to $$ \frac{\partial F}{\partial x} = -3yz $$ $$ \frac{\partial F}{\partial y} = -3xz $$ $$ \frac{\partial F}{\partial z} = 3z^2-3xy $$ Hence $$ \frac{\partial z}{\partial x} = -\frac{F_x}{F_z}= -\frac{-3yz}{3z^2-3xy} $$ and $$ \frac{\partial z}{\partial y} = -\frac{F_y}{F_z}= -\frac{-3xz}{3z^2-3xy} $$