Let's determine the value of the product $x y$, given that $x$ and $y$ are distinct numbers such that

$$
\frac{1}{1+x^{2}}+\frac{1}{1+y^{2}}=\frac{2}{1+x y}
$$
Multiplying both sides by the product of the denominators, we get

$$
\left(1+y^{2}\right)(1+x y)+\left(1+x^{2}\right)(1+x y)=2\left(1+x^{2}\right)\left(1+y^{2}\right)
$$

After performing the operations and rearranging, we have

$$
x^{2}+y^{2}+2 x^{2} y^{2}-x^{3} y-x y^{3}-2 x y=0
$$

The left side can be factored as:

$$
x^{2}+y^{2}+2 x^{2} y^{2}-x^{3} y-x y^{3}-2 x y=(x-y)^{2}-x y(x-y)^{2}=(x-y)^{2}(1-x y)
$$

A product is zero if and only if one of its factors is zero. According to the condition, the first factor cannot be 0, so $1-x y=0$.

Thus, we have obtained that under the given conditions, $x y=1$. In this case, the denominator of the fractions involved is not zero. The problem is solved.