Answer
$-1$ and $-1$
Work Step by Step
Consider $F(x,y,z)=\sin (x+y)+\sin (y+z)+\sin (x+z)$
and $\dfrac{\partial z}{ \partial x}=-\dfrac{F_x}{F_z}$
$\implies \dfrac{\partial z}{ \partial x}=-[\dfrac{\cos (x+y)+\cos (x+z)}{\cos (y+z)+\cos (x+z)}]$
For the point $(\pi,\pi,\pi)$, we have
$\dfrac{\partial z}{ \partial x}=-[\dfrac{\cos (\pi+\pi)+\cos (\pi+\pi)}{\cos (\pi+\pi)+\cos (\pi+\pi)}]$
or, $=-1$
Now, consider $\dfrac{\partial z}{ \partial y}=-\dfrac{F_y}{F_z}$ .
$\implies \dfrac{\partial z}{ \partial y}=-[\dfrac{\cos (x+y)+\cos (y+z)}{\cos (y+z)+\cos (x+z)}]$
For the point $(\pi,\pi,\pi)$, we have: $\dfrac{\partial z}{ \partial y}(\pi,\pi,\pi)=-[\dfrac{\cos (\pi+\pi)+\cos (\pi+\pi)}{\cos (\pi+\pi)+\cos (\pi+\pi)}]=-1$
