20.2. (NBR, 68). Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$, satisfying the identity

$$
x f(y) + y f(x) \equiv (x + y) f(x) f(y), \quad x, y \in \mathbb{R}
$$
20.2. Let's set \( x = y = 1 \) in the original identity, then we get \( 2 f(1) = 2 (f(1))^2 \), i.e., \( f(1) = 0 \) or \( f(1) = 1 \). Let's consider each of these cases:

a) If \( f(1) = 0 \), then by setting \( y = 1 \) in the identity, we get the identity \( f(x) \equiv 0 \).

b) If \( f(1) = 1 \), then by setting \( y = 1 \) again, we get the identity

\[
f(x) + x \equiv (x + 1) f(x), \text{ i.e., } x(f(x) - 1) \equiv 0
\]

from which for any \( x \neq 0 \) we have \( f(x) = 1 \). Thus, for the function \( f(x) \) there are two possibilities: either \( f(x) \equiv 0 \), or

\[
f(x) = \begin{cases}
1 & \text{if} & x \neq 0, \\
a & \text{if} & x = 0, \text{ where } a \in \mathbb{R}.
\end{cases}
\]

Verification shows that any of these functions satisfies the condition of the problem.