$\frac{\partial{z}}{\partial{y}}$ if $z$ is given implicitly

Question

Suppose $z$ is given implicitly as:

$$e^z-x^2y-y^2z = 0$$

Find $$\frac{\partial{z}}{\partial{y}}$$.

I let $F(x,y) = e^z-x^2y-y^2z$. Then,

$$\frac{\partial{F}}{\partial{y}} = -x^2-2yz$$ $$\frac{\partial{F}}{\partial{z}} = e^z-y^2$$

$$\frac{\frac{\partial{F}}{\partial{y}}}{\frac{\partial{F}}{\partial{z}}} = \frac{\partial{F}}{\partial{y}} \cdot \frac{\partial{z}}{\partial{F}} = \frac{\partial{z}}{\partial{y}}$$

Therefore,

$$\frac{\partial{z}}{\partial{y}} = \frac{-x^2-2yz}{e^z-y^2}$$

However, the exercise cites the answer as

$$\frac{x^2+2yz}{e^z-y^2}$$

For some reason, I am not seeing a way to remove the negative, or the given answer is a mistake.

Answer 1

Looking through my copy of Stewart's calculus we have

$$\frac{\partial z}{\partial x} = -\frac{\frac{\partial F}{\partial x}}{\frac{\partial F}{\partial z}}$$

and similarly

$$\frac{\partial z}{\partial y} = -\frac{\frac{\partial F}{\partial y}}{\frac{\partial F}{\partial z}}$$

So basically you just have your formula written down wrong. So you are only a negative sign off. Everything else is perfect.

Answer 2

Maybe the apparent proximity to the given result is misleading. Treat $z$ as an implicit function.

$e^z-x^2y-y^2z = 0$

$z_ye^z-x^2-2yz-y^2z_y=0$

$z_y(e^z-y^2)=x^2+2yz$

Leading to the given answer: $z_y=\dfrac{x^2+2yz}{e^z-y^2}$

Added

This proves the minus sign in the formula you used.