, 2020

Denote the original relation by (*), and define $P(a, b)$ as the property obtained by substituting $x = a, y = b$ into the given condition on the function $f$. We analyze this through the following steps:

(A). Applying $P(x, 1)$ gives $f(x + f(1)) = f(1) + f(f(x))$. From this we obtain $f(f(1 - f(1))) = 0$. Organizing this as follows:
$$f(1 - f(1)) = a, \quad f(a) = 0. \tag{1}$$
Also, applying $P(a, a)$ and using (1) gives
$$f(a^2) = f(0). \tag{2}$$
Then applying $P(0, a^2)$ and $P(0, x)$, we can successively derive
$$f(0) = 0, \quad f(f(x)) = f(x), \quad \forall x \in \mathbb{R}. \tag{3}$$

(B). Suppose there exists $t \neq 0, 1$ such that $f(t) = 0$. Then from $P(x, t)$ and (3) we obtain $f(xt) = t f(x), \forall x \in \mathbb{R}$. Then from $P(tx, ty) - t P(x, y)$ we derive $(t^2 - t)(f(xy) - y f(x)) = 0$, and setting $x = 1$ gives $f(y) = f(1) y$, which combined with condition (*) gives $f(1) = 0$ or $f(1) = 1$. Thus we obtain two possible functions: $f(y) = 0$ or $f(y) = y, \forall y \in \mathbb{R}$. Clearly $f(y) = 0$ satisfies condition (*), while $f(y) = y$ does not satisfy the condition in this case. (Because $f(t) \neq 0, \forall t \neq 0$.)

(C). Excluding the case in (B), we now assume $f(t) \neq 0, \forall t \in \mathbb{R} - \{0, 1\}$. Then (1) gives $f(1) = 0$ or $f(1) = 1$. If $f(1) = 0$, then from $P(1, y)$ we obtain $f(1 + f(y)) = 0$. Therefore, $f(y) = -1, \forall y \in \mathbb{R} - \{0, 1\}$, but this clearly contradicts (*). Hence $f(1) = 1$, and we have the following results:
$$f(0) = 0, \quad f(1) = 1; \quad f(t) \neq 0, \forall t \neq 0. \tag{4}$$

From $P(x, 1)$ we obtain $f(x+1) = f(x)+1$. Then from $P(x, y+1)-P(x, y)$ we obtain $f(x+xy) = f(x) + f(xy)$. Therefore, when $x \neq 0$, for any real number $z$ we may set $y = \frac{z}{x}$ to obtain
$$f(x+z) = f(x) + f(z). \tag{5}$$
Also noting that $f(0) = 0$, this result clearly holds for all real numbers $x, z$. Then from $P(x, 1)$ we can further obtain $f(f(x)) = f(x)$. From this result together with applying the property (5) into (*), we deduce that $f(xy) = y f(x)$. Finally, from this equation combined with the argument in (B), we must obtain $f(y) = y, \forall y \in \mathbb{R}$. Clearly this equation satisfies the given condition.

In summary, the functions satisfying the condition are $f(x) = 0, \forall x \in \mathbb{R}$ and $f(x) = x, \forall x \in \mathbb{R}$.