We are tasked with finding all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ satisfying the equation:

$$f(xy) = f(x)f(y) + f(f(x+y))$$

for all $x, y \in \mathbb{R}$.

### Step 1: Consideration of Simple Cases

First, let us consider the case where $x = 0$:

$$f(0) = f(0)f(y) + f(f(y))$$

for all $y \in \mathbb{R}$. Let $f(0) = c$. Thus, we have:

$$c = cf(y) + f(f(y))$$

Rearranging gives us:

$$f(f(y)) = c(1 - f(y))$$

### Step 2: Examine Constant Solutions

Let us consider if a constant function $f(x) = c$ could be a solution:

1. If $f(x) = 0$ for all $x$, then the equation becomes $0 = 0 + 0$, which is always true. Hence, $f(x) = 0$ is indeed a solution.

2. For nonzero constant $f(x) = c$, substitute into the original equation:

$$c = c \cdot c + c$$

which implies:

$$c = c^2 + c$$

leading to the quadratic equation:

$$c^2 + c - c = 0, \quad \Rightarrow \quad c(c+1) = 0$$

This gives $c = 0$ or $c = -1$. Thus, we need to check if $f(x) = -1$ is a valid solution. Substituting back:

$$f(xy) = -1 = (-1)(-1) + f(f(x+y)) = 1 + f(f(x+y))$$

Then $f(f(x+y)) = -2$, which contradicts with the hypothesis that $f \equiv -1$. So $f(x) = -1$ is not a valid solution.

### Step 3: Consider Linear Solutions

Let's test if a linear function satisfies the equation. Suppose $f(x) = ax + b$. Substitute into the original equation:

$$f(xy) = a(xy) + b = (ax + b)(ay + b) + f(a(x+y) + b)$$

which simplifies to:

$$axy + b = a^2xy + abx + aby + b^2 + f(ax + ay + 2b)$$

To satisfy this for all $x, y$, we must have coefficients of $xy$: $a = a^2 \Rightarrow a(a - 1) = 0$. Hence, $a = 0$ or $a = 1$.

- If $a = 0$, $f(x) = b = 0$ (since we already know $f(x) = 0$ is a solution).

- If $a = 1$, substitute into simplified form:

$$xy + b = xy + x + y + b^2 + f(x + y + b)$$

From this, consequently $b = -1$. Thus, $f(x) = x - 1$.

### Conclusion

The solutions are:

- $f(x) = 0$
- $f(x) = x - 1$

Thus, the functions $f(x) = 0$ and $f(x) = x - 1$ satisfy the given functional equation, and these are the solutions. Therefore, the final answer is:

$$\boxed{\text{f(x) = 0 and f(x) = x - 1}}$$