Answer
$-1,-1$
Work Step by Step
Since, we have $F(x,y,z)=\sin (x+y)+\sin (y+z)+\sin (x+z)$
$\dfrac{\partial z}{ \partial x}=-\dfrac{F_x}{F_z}$ ...(1)
Now, need to plug the derivatives in equation (1).
we have $\dfrac{\partial z}{ \partial x}=-\dfrac{\cos (x+y)+\cos (x+z)}{\cos (y+z)+\cos (x+z)}$
Now, plug the point $(\pi,\pi,\pi)$
Thus, we get $\dfrac{\partial z}{ \partial x}=-\dfrac{\cos (\pi+\pi)+\cos (\pi+\pi)}{\cos (\pi+\pi)+\cos (\pi+\pi)}=-1$
Now, $\dfrac{\partial z}{ \partial y}=-\dfrac{F_y}{F_z}$ ...(2)
Plug the derivatives in equation (2).
we have $\dfrac{\partial z}{ \partial y}=-\dfrac{\cos (x+y)+\cos (y+z)}{\cos (y+z)+\cos (x+z)}$
and $\dfrac{\partial z}{ \partial y}(\pi,\pi,\pi)=-\dfrac{\cos (\pi+\pi)+\cos (\pi+\pi)}{\cos (\pi+\pi)+\cos (\pi+\pi)}=-1$