For $x = 0$ in the given relation we get $f(f(0)) + y f(0) \le 0$, for every $y \in \mathbb{R}$, which is true only if $f(0) = 0$. Indeed, if $f(0) \ne 0$, then $y \le \frac{-f(f(0))}{f(0)}$, for every $y \in \mathbb{R}$, or $y \ge \frac{-f(f(0))}{f(0)}$, for every $y \in \mathbb{R}$, which is impossible.

Choosing $y = 0$, it results $f(f(x)) \le x$, for every real $x$. (1)

For $x = 1$ in the given relation we get $f(f(1)) + y f(1) \le 1 + f(f(y))$, and using (1) we get $f(f(1)) + y f(1) \le 1 + y$, so $y(f(1) - 1) \le 1 - f(f(1))$, for every real $y$. The relation is true, as above, only if $f(1) - 1 = 0$, i.e. $f(1) = 1$.

In the above context, for $x = 1$, the relation from the hypothesis ensures that $y \le f(f(y))$, for every real $y$. (2)

From (1) and (2) we get that $f(f(x)) = x$ for every $x \in \mathbb{R}$, and using the hypothesis we obtain that $x + y f(x) \le x + x y$, so $y f(x) \le x y$ for every $x, y \in \mathbb{R}$.

For $y = 1$ we get $f(x) \le x$ and for $y = -1$ we get $f(x) \ge x$, so $f(x) = x$ for every $x \in \mathbb{R}$, which satisfies the hypothesis.