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

To solve this I first rearrange $$z^3 - 3xyz - 1 = 0$$

$$\displaystyle\frac{\partial z}{\partial x}$$ = $$3z^2z'-3xyz'-3yz$$

$$\displaystyle\frac{\partial z}{\partial y}$$ = $$3zz' -3xyz'-3z$$

$$\displaystyle\frac{\partial z}{\partial z}$$ = $$3z^2-3xy$$

Now to apply the implicit function theorem do I just need to plug in the values in the formula $$\displaystyle\frac{\partial z}{\partial x} = \frac{-F_x}{F_z}$$ and $$\displaystyle\frac{\partial z}{\partial y} = \frac{-F_y}{F_z}$$ ?

I would require some directions solving this if I am wrong.

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 $$\frac{\partial z}{\partial y} = -\frac{F_y}{F_z}= -\frac{-3xz}{3z^2-3xy}$$