To solve the functional equation for $f : \mathbb{R} \to \mathbb{R}$,

$$f(x)f(y) = xf(f(y-x)) + xf(2x) + f(x^2),$$

for all real $x, y$, we proceed as follows:

1. **Substitute $y = 0$:**
Considering $y = 0$, the equation becomes:
$$f(x)f(0) = xf(f(-x)) + xf(2x) + f(x^2).$$
Notice that if $f$ is a constant zero function, $f(x) = 0$ for all $x$, then the equation holds trivially. Hence, $f(x) = 0$ is a solution.

2. Check for non-zero solutions:
Assume there exists $x_0$ such that $f(x_0) \neq 0$. Then $f(y)$ cannot be zero for all $y$. With this assumption, let's explore further.

3. **Substitute $x = 0$:**
$$f(0)f(y) = 0 \quad \text{for all } y.$$
This implies $f(0) = 0$ or $f(y) = 0$ for all $y$. Since we're considering non-zero solutions, we assume $f(0) = 0$.

4. Substitute specific values to find a pattern:

Let's use simple substitutions to analyze behavior at specific points.

**Substitute $x = 1$:**
$$f(1)f(y) = f(f(y-1)) + f(2) + f(1).$$

This implies that if $f(1)$ is known, we could potentially express $f(y)$ in simpler terms.

5. Try a linear solution:
Assume $f(x) = cx$ for some constant $c$. Substitute into the original equation:
$$(cx)(cy) = x(c(c(y-x))) + x(c(2x)) + c(x^2).$$
Simplifying,
$$c^2xy = x(c^2(y-x)) + 2cx^2 + cx^2.$$
$$c^2xy = c^2xy - c^2x^2 + 3cx^2.$$
Equating both sides, we get:
$$0 = -c^2x^2 + 3cx^2.$$
$$x^2(c^2 - 3c) = 0.$$
Which implies:
$$c^2 - 3c = 0 \quad \Rightarrow \quad c(c-3) = 0.$$
Thus, $c = 0$ or $c = 3$, leading to the solutions:
$$f(x) = 0 \quad \text{or} \quad f(x) = 3x.$$

Thus, the functions that satisfy the given functional equation are $f(x) = 0$ and $f(x) = 3x$ for all $x \in \mathbb{R}$.

$$\boxed{f(x) = 0 \quad \text{and} \quad f(x) = 3x \quad \text{for all } x \in \mathbb{R}}$$