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