Answer
$-9,-4$
Work Step by Step
Since, we have $F(x,y,z)= z^3-xy+yz+y^3-2=0$
$\dfrac{\partial z}{ \partial x}=-\dfrac{F_x}{F_z}$ ...(1)
Now, need to plug the derivatives in equation (1).
we have $\dfrac{\partial z}{ \partial x}=-\dfrac{-(1/x^2)}{(-1/z^2)}=-\dfrac{z^2}{x^2}$
Now, plug the point $(2,3,6)$
Thus, we get $\dfrac{\partial z}{ \partial x}=-\dfrac{6^2}{2^2}=-9$
Now, $\dfrac{\partial z}{ \partial y}=-\dfrac{F_y}{F_z}$ ...(2)
Plug the derivatives in equation (2).
we have $\dfrac{\partial z}{ \partial y}=-\dfrac{-(1/y^2)}{(-1/z^2)}=-\dfrac{z^2}{y^2}$
and $\dfrac{\partial z}{ \partial y}(2,3,6)=-\dfrac{6^2}{3^2}=-4$
