Answer
\begin{aligned} \frac{\partial z }{\partial x}=-\sec (y+z) \end{aligned}
\begin{aligned} \frac{\partial z }{\partial y}=-1 \end{aligned}
Work Step by Step
Given $$x+ \sin(y+z)=0$$
by letting $$F(x,y,z)= x+ \sin(y+z)$$
So, we have
$$F_x(x,y,z)=\frac{\partial F(x,y,z)}{\partial x}=1\\
$$
$$F_y(x,y,z)=\frac{\partial F(x,y,z)}{\partial y}= \cos(y+z)$$
$$F_z(x,y,z)=\frac{\partial F(x,y,z)}{\partial z}=\cos(y+z)$$
Also, we get
\begin{aligned} \frac{\partial z }{\partial x}&=-\frac{F_{x}(x, y,z)}{F_{z}(x, y,z)} \\
&=- \frac{1}{\cos(y+z)}\\ &=-\sec (y+z) \end{aligned}
\begin{aligned} \frac{\partial z }{\partial y}&=-\frac{F_{y}(x, y,z)}{F_{z}(x, y,z)} \\
&=- \frac{\cos(y+z)}{\cos(y+z)} \\ &=-1 \end{aligned}
