Answer
\begin{aligned} \frac{\partial z }{\partial x} =- \frac{\sec^2(x+y)}{\sec^2(y+z)} \end{aligned}
\begin{aligned}
\frac{\partial z }{\partial y} =-\left(\frac{ \sec^2(x+y)}{\sec^2(y+z)}+1\right) \end{aligned}
Work Step by Step
Given $$\tan(x+y)+ \tan(y+z)=1$$
$$ \Rightarrow \tan(x+y)+ \tan(y+z)-1=0$$
by letting $$F(x,y,z)= \tan(x+y)+ \tan(y+z)-1$$
So, we have
$$F_x(x,y,z)=\frac{\partial F(x,y,z)}{\partial x}=\sec^2(x+y)\\
$$
$$F_y(x,y,z)=\frac{\partial F(x,y,z)}{\partial y}= \sec^2(x+y)+\sec^2(y+z)$$
$$F_z(x,y,z)=\frac{\partial F(x,y,z)}{\partial z}=\sec^2(y+z)$$
Also, we get
\begin{aligned} \frac{\partial z }{\partial x}&=-\frac{F_{x}(x, y,z)}{F_{z}(x, y,z)} \\
&=- \frac{\sec^2(x+y)}{\sec^2(y+z)} \end{aligned}
\begin{aligned}
\frac{\partial z }{\partial y}&=-\frac{F_{y}(x, y,z)}{F_{z}(x, y,z)} \\
&=- \frac{\sec^2(x+y)+\sec^2(y+z)}{\sec^2(y+z)} \\ &=-\left(\frac{ \sec^2(x+y)}{\sec^2(y+z)}+1\right) \end{aligned}
