Middle European Mathematical Olympiad

To solve the functional equation problem, we need to identify all functions $f: \mathbb{R} \to \mathbb{R}$ that satisfy the given functional equation for any real numbers $x, y$:

$$f(xf(y) + 2y) = f(xy) + xf(y) + f(f(y)).$$

We will explore potential solutions by substituting specific values for $x$ and $y$, as well as considering general assumptions.

### Step 1: Testing for Constant Solutions

Suppose $f$ is a constant function, i.e., $f(x) = c$ for all $x \in \mathbb{R}$. Substituting into the equation gives:

$$f(xf(y) + 2y) = f(xy) + xf(y) + f(f(y)).$$

This simplifies to:

$$c = c + xc + c.$$

Thus, for constant solutions, we see:

$$xc = 0.$$

Since this must hold for all $x \in \mathbb{R}$, it implies $c = 0$. Therefore, $f(x) = 0$ for all $x$ is a solution.

### Step 2: Checking for Linear Solutions

Suppose $f(x) = ax$ where $a$ is a constant. Substituting into the functional equation gives:

$$f(xf(y) + 2y) = ax(ay) + 2ay = a^2xy + 2ay,$$

$$f(xy) + xf(y) + f(f(y)) = a(xy) + x(ay) + a(ay) = a(xy) + axy + a^2y.$$

Equating both expressions:

$$a^2xy + 2ay = a(xy) + axy + a^2y.$$

This simplifies to:

$$a^2xy + 2ay = 2axy + a^2y.$$

By equating coefficients, the terms $a^2xy$ and $2axy$ suggest $a^2 = 2a$, leading to $a(a - 2) = 0$. Thus, $a = 0$ or $a = 2$.

- If $a = 0$, then $f(x) = 0$ for all $x$, which we have already identified.
- If $a = 2$, then $f(x) = 2x$.

Therefore, $f(x) = 2x$ satisfies the functional equation as well.

### Conclusion

We have found two solutions to the functional equation:

1. $f(x) = 0$ for all $x \in \mathbb{R}$.
2. $f(x) = 2x$ for all $x \in \mathbb{R}$.

So, the set of functions satisfying the equation is

$$\boxed{f(x) = 0 \quad \text{or} \quad f(x) = 2x}.$$

These are the only functions that satisfy the given functional equation.