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.


1 Answer 1


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


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