Answer
\begin{aligned} \frac{\partial w }{\partial x} =\frac{y+w}{z-x} \\
\end{aligned}
\begin{aligned} \frac{\partial w }{\partial y} =\frac{x+z}{z-x} \\
\end{aligned}
\begin{aligned} \frac{\partial w }{\partial z} =\frac{y-w}{z-x} \\
\end{aligned}
Work Step by Step
Given $$x y+y z-w z+w x=5$$
$$\Rightarrow x y+y z-w z+w x-5=0$$
by letting $$F(x,y,z,w)= x y+y z-w z+w x-5$$
So, we have
$$F_x(x,y,z,w)=\frac{\partial F(x,y,z,w)}{\partial x}= y+w\\
$$
$$F_y(x,y,z,w)=\frac{\partial F(x,y,z,w)}{\partial y}= x+z$$
$$F_z(x,y,z,w)=\frac{\partial F(x,y,z,w)}{\partial z}=y-w$$
$$F_w(x,y,z,w)=\frac{\partial F(x,y,z,w)}{\partial w}=-z+x$$
Also, we get
\begin{aligned} \frac{\partial w }{\partial x}&=-\frac{F_{x}(x, y,z,w)}{F_{w}(x, y,z,w)} \\
&=-\frac{y+w}{-z+x} \\
&=\frac{y+w}{z-x} \\
\end{aligned}
\begin{aligned} \frac{\partial w }{\partial y}&=-\frac{F_{y}(x, y,z,w)}{F_{w}(x, y,z,w)} \\
&=-\frac{x+z}{-z+x} \\
&=\frac{x+z}{z-x} \\
\end{aligned}\begin{aligned} \frac{\partial w }{\partial z}&=-\frac{F_{z}(x, y,z,w)}{F_{w}(x, y,z,w)} \\
&=-\frac{y-w}{-z+x} \\
&=\frac{y-w}{z-x} \\
\end{aligned}
