Partial Derivative of implicit function z defined as a function of x and y

Question

Suppose that z is defined implicitly as a function of x and y by the equation

$ x^2 + yz - z^3 = 0$

Calculate the partial derivatives $\frac{\partial z}{\partial y} and \frac{\partial z}{\partial x}$ at (x,y) = (1,0). Answer should be numbers that don't depend on any variables

I know that I must take the derivative with respect to x and then y, but I always have a lingering $z^2$ laying around that I don't know how to get rid of. For example derivative w.r.t x:

$2x + y\frac{\partial z}{\partial x} - 3z^2\frac{\partial z}{\partial x }=0$

Answer 1

This question just wants you to find the partial derivatives of x and y with respect to z. So start with $F_x$. This is clearly $2x+y\frac{∂z}{∂x}-3z^2 \frac{∂z}{∂x}=0$. Then compute $F_y$, which is $\frac{∂z}{∂y}-3z^2\frac{∂z}{∂y}=0$. To solve this you're going to need to know the value of z at (1,0). This is clearly 1. Now plug the set of points (1,0,1) into your partial derivatives.$F_x$= $2(1)-3(1)^2\frac{∂z}{∂x}=0 $. Here you find $\frac{∂z}{∂x} = 2/3$. Computing similarly for $\frac{∂z}{∂y}$ you get 0.

As I showed, with the lingering $3z^2$ you plug in the value for z you determine by using the initial equation.

Answer 2

For the case of finding $\frac{\partial z}{\partial x}(1,0)$ just differentiate the equation w.r.t $x$ to give $$2x + y \frac{\partial z}{\partial x} - 3z^2 \frac{\partial z}{\partial x} = 0.$$

Next substitute the point $(x,y) = (1,0)$ into this expression. Then use the original equation to evaluate $z(x = 1,y = 0)$ and substitute this into the above expression. This will leave a numeric value for $\frac{\partial z}{\partial x}(1,0)$ as required.

Apply an analogous process to calculate $\frac{\partial z}{\partial y}(1,0)$.