Answer
$\dfrac{1}{4}$ and $\dfrac{-3}{4}$
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 $\dfrac{\partial z}{ \partial x}=-\dfrac{-y}{3z^2+y}=\dfrac{y}{3z^2+y}$
Now, plug the point $(1,1,1)$
Thus, we get $\dfrac{\partial z}{ \partial x}=\dfrac{1}{4}$
$\dfrac{\partial z}{ \partial y}=-\dfrac{F_y}{F_z}$ ...(1)
Now, need to plug the derivatives in equation (1).
we have $\dfrac{\partial z}{ \partial y}=\dfrac{x-z-3y^2}{3z^2+y}$
and $\dfrac{\partial z}{ \partial y}(1,1,1)=\dfrac{1-1-3}{3+1}=\dfrac{-3}{4}$
