## Thomas' Calculus 13th Edition

$-9$ and $-4$
Consider $\dfrac{\partial z}{ \partial x}=-\dfrac{F_x}{F_z}$ $\implies \dfrac{\partial z}{ \partial x}=-\dfrac{-(1/x^2)}{(-1/z^2)}=-(\dfrac{z^2}{x^2})$ For the point $(2,3,6)$, we have Now, $\dfrac{\partial z}{ \partial x}=-\dfrac{6^2}{2^2}=-9$ and, $\dfrac{\partial z}{ \partial y}=-\dfrac{F_y}{F_z}$ $\implies \dfrac{\partial z}{ \partial y}=-\dfrac{-(1/y^2)}{(-1/z^2)}=-(\dfrac{z^2}{y^2})$ Also, $\dfrac{\partial z}{ \partial y}(2,3,6)=-\dfrac{36}{9}=-4$