Problem 2. Determine the value of $\frac{1}{x^{2}+1}+\frac{1}{y^{2}+1}+\frac{2}{x y+1}$, if $x \neq y$ and $\frac{1}{x^{2}+1}+\frac{1}{y^{2}+1}=\frac{2}{x y+1}$.
Solution. The given equation can be written in the form

$$
\begin{aligned}
& \frac{1}{x^{2}+1}-\frac{1}{x y+1}=\frac{1}{x y+1}-\frac{1}{y^{2}+1} \\
& \frac{x(y-x)}{\left(x^{2}+1\right)(x y+1)}=\frac{y(y-x)}{\left(y^{2}+1\right)(x y+1)}
\end{aligned}
$$

Since $x \neq y$ and $x y+1 \neq 0$, we obtain the equation $\frac{x}{x^{2}+1}=\frac{y}{y^{2}+1}$, which is equivalent to the equation:

$$
(x y-1)(x-y)=0
$$

From $x-y \neq 0$, we get $x y-1=0$, i.e., $x y=1$. Now

$$
\frac{1}{x^{2}+1}+\frac{1}{y^{2}+1}+\frac{2}{x y+1}=\frac{2}{x y+1}+\frac{2}{x y+1}=\frac{4}{1+1}=2
$$