Find $\frac{\partial z}{\partial x}$ and $\frac{\partial z}{\partial y}$ given $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}-1=0$ at $P=(2,3,6)$

Question

I am given the following implicit differentiation problem to work on:

Find $\dfrac{\partial z}{\partial x}$ and $\dfrac{\partial z}{\partial y}$ given $\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}-1=0$ at $P=(2,3,6)$

The textbook contains no answer so I would like to check my work, please.

Using the implicit differentiation theorem we have that

$$F(x,y,z) = \frac{1}{x}+\frac{1}{y}+\frac{1}{z}-1\\ F_x(x,y,z) = - \frac{1}{x^2}, \quad F_y(x,y,z) = - \frac{1}{y^2}$$

Therefore we have

$$\frac{\partial z}{\partial x} (P) = - \frac{F_x}{F_z} = - \frac{z^2}{x^2} = -9, \quad \frac{\partial z}{\partial y} (P) = - \frac{F_y}{F_z} = - \frac{z^2}{y^2} = -4$$

Any input and/or tip in case I solved it wrong is highly appreciated,

Thank you

Answer 1

It is correct, though you could directly implicitly differentiate: $$\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}-1\right)'_x=0 \Rightarrow -\frac{1}{x^2}-\frac{1}{z^2}\cdot z_x=0 \Rightarrow z_x=-\frac{z^2}{x^2}.$$