To find all functions $f : \mathbb{R} \to \mathbb{R}$ such that $f(0) \neq 0$ and

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

for all $x, y \in \mathbb{R}$, we proceed as follows.

- Step 1: Initial Substitution

Start by substituting $x = 0$ and $y = 0$ into the original functional equation:

$$f(f(0)) + f(f(0)) = f(0 + 0) f(0 \cdot 0).$$

This simplifies to:

$$2f(f(0)) = f(0)^2.$$

Since $f(0) \neq 0$, we can solve this to find that

$$f(f(0)) = \frac{f(0)^2}{2}.$$

- Step 2: Explore Further Substitution

Let $y = 0$ and rewrite the equation:

$$f(f(x)) + f(f(0)) = f(x)f(0).$$

Substitute the value from Step 1:

$$f(f(x)) + \frac{f(0)^2}{2} = f(x)f(0).$$

Rearrange to obtain:

$$f(f(x)) = f(x)f(0) - \frac{f(0)^2}{2}.$$

This equation will be useful in confirming specific solutions.

- Step 3: Consider a Constant Function Solution

Evaluate the possibility of constant solutions by letting $f(x) = c$ for some constant $c$.

Substitute $f(x) = c$ into the functional equation:

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

This simplifies to:

$$2c = c^2.$$

Solve this quadratic equation:

$$c^2 - 2c = 0,$$

which factors to:

$$c(c - 2) = 0.$$

Therefore, $c = 0$ or $c = 2$. Since $f(0) \neq 0$, $c = 0$ is not valid, leaving $c = 2$.

Thus, $f(x) = 2$ for all $x \in \mathbb{R}$ is a potential solution.

- Step 4: Verification

Verify that $f(x) = 2$ satisfies the functional equation:

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

Substituting $f(x) = 2$ yields:

$$2 + 2 = 2 \cdot 2.$$

Which simplifies to $4 = 4$, confirming it is a solution.

Thus, the only function satisfying all the conditions is the constant function:
$$\boxed{f(x) = 2 \quad \forall x \in \mathbb{R}}.$$