Answer
\begin{aligned} \frac{\partial w }{\partial x} =\frac{2x}{5y-20w} \\
\end{aligned}
\begin{aligned} \frac{\partial w }{\partial y} =\frac{2y-5w}{5y-20w} \\
\end{aligned}
\begin{aligned} \frac{\partial w }{\partial z} =\frac{2z}{5y-20w} \\
\end{aligned}
Work Step by Step
Given $$x^{2}+y^{2}+z^{2}-5 y w+10 w^{2}=2$$
$$\Rightarrow x^{2}+y^{2}+z^{2}-5 y w+10 w^{2}-2=0$$
by letting $$F(x,y,z,w)= x^{2}+y^{2}+z^{2}-5 y w+10 w^{2}-2$$
So, we have
$$F_x(x,y,z,w)=\frac{\partial F(x,y,z,w)}{\partial x}= 2x\\
$$
$$F_y(x,y,z,w)=\frac{\partial F(x,y,z,w)}{\partial y}= 2y-5w$$
$$F_z(x,y,z,w)=\frac{\partial F(x,y,z,w)}{\partial z}=2z$$
$$F_w(x,y,z,w)=\frac{\partial F(x,y,z,w)}{\partial w}=-5y+20w$$
Also, we get
\begin{aligned} \frac{\partial w }{\partial x}&=-\frac{F_{x}(x, y,z,w)}{F_{w}(x, y,z,w)} \\
&=-\frac{2x}{-5y+20w} \\
&=\frac{2x}{5y-20w} \\
\end{aligned}
\begin{aligned} \frac{\partial w }{\partial y}&=-\frac{F_{y}(x, y,z,w)}{F_{w}(x, y,z,w)} \\
&=-\frac{2y-5w}{-5y+20w} \\
&=\frac{2y-5w}{5y-20w} \\
\end{aligned}\begin{aligned} \frac{\partial w }{\partial z}&=-\frac{F_{z}(x, y,z,w)}{F_{w}(x, y,z,w)} \\
&=-\frac{2z}{-5y+20w} \\
&=\frac{2z}{5y-20w} \\
\end{aligned}
