#### Answer

$$\frac{\partial w}{\partial y}\bigg|_{(1,1,1)} =\frac{-1}{2}$$

#### Work Step by Step

Given $$\frac{1}{w^{2}+x^{2}}+\frac{1}{w^{2}+y^{2}}=1$$
Consider
$$F(x,y,w)=\frac{1}{w^{2}+x^{2}}+\frac{1}{w^{2}+y^{2}}=1$$
Since
\begin{align*}
F_{y}&=-\frac{2 y}{\left(w^{2}+y^{2}\right)^{2} } \\
F_{w}&=\frac{-2 w}{\left(w^{2}+x^{2}\right)^{2}}-\frac{2 w}{\left(w^{2}+y^{2}\right)^{2}}
\end{align*}
Then
\begin{align*}
\frac{\partial w}{\partial y}&=-\frac{F_y}{F_w}\\
&=-\frac{\frac{-2 y}{\left(w^{2}+y^{2}\right)^{2}}}{\frac{-2 w}{\left(w^{2}+x^{2}\right)^{2}}-\frac{2 w}{\left(w^{2}+y^{2}\right)^{2}}}\\
&=-\frac{y\left(w^{2}+x^{2}\right)^{2}}{w\left(w^{2}+y^{2}\right)^{2}+w\left(w^{2}+x^{2}\right)^{2}}\\
&=\frac{-y\left(w^{2}+x^{2}\right)^{2}}{w\left(\left(w^{2}+y^{2}\right)^{2}+\left(w^{2}+x^{2}\right)^{2}\right)}
\end{align*}
Hence
\begin{align*}
\frac{\partial w}{\partial y}\bigg|_{(1,1,1)}&=\frac{-1}{2}
\end{align*}