Here's the question:
Given the equation $z^3-xz-y=0$.
Prove that the equation defines an implicit function $z(x,y)$ in a neighborhood of point $(1,0,-1)$.
Calculate $\frac{\partial ^2 z}{\partial x \partial y}$ in point $(1,0)$
Solution:
I defined $F(x,y,z)=z^3-xz-y$
We have that $F(1,0,-1)=0$
The function $F$ is differentiable and it's partial derivatives are differentiable (polynomial).
$F'_z(1,0,-1)=3(-1)^2-1=2\ne 0$
And thus, there exists a function $z(x,y)$ in a neighborhood of $(1,0,-1)$.
And we know that:
$Z'_y=- \frac{F'y}{F'z}=- \frac{-1}{3z^2-x}=\frac{1}{3z^2-x}$
And so $\frac{\partial ^2 z}{\partial x \partial y}(1,0,-1)=\frac{1}{(3z^2-x)^2}=\frac{1}{(3(-1)^2-1)^2}=\frac{1}{4}$
I am mostly suspicious of my last calculation of the derivative, I am not sure of why I did the point $(1,0,-1)$, I would love to know why is it correct or false and any general feedback about my solution.
Thanks in advance.
Fixing derivative with help of Joe:
$\frac{\partial ^2 z}{\partial x \partial y}=\frac{-(6z*(\frac{-z}{3z^2-x})-1)}{(3z^2-x)^2}=\frac{1+\frac{6z^2}{3z^2-x}}{(3z^2-x)^2}$
And in $(1,0,-1)$---> $=1$
I get the same answer as you, except that I have:
$F_x = -z$
$F_z = 3z^2 - x$
$\frac{\partial z}{\partial x} = -\frac{F_x}{F_z} = \frac{z}{3z^2-x}$
So for the final answer I get: $\frac{-(6z*(\frac{z}{3z^2-x})-1)}{(3z^2-x)^2}=\frac{1-\frac{6z^2}{3z^2-x}}{(3z^2-x)^2} = \frac{1-3}{4}= \frac{-1}{2}$
The reason that they asked for the partial derivative at $(1,0)$, instead of at $(1,0,-1)$, is because they were asking for the partial derivative of the function $z(x,y)$. However, you can still substitute $-1$ in for $z$ in the final answer, since you know that $z(1,0)=-1$.