Let $f : \mathbb{R} \rightarrow \mathbb{R}$ be a function satisfying the functional equation:

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

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

We aim to find all such functions $f$.

### Step 1: Substitution and Simplification

First, substitute $y = 0$ into the given equation:

$$(f(x) + 0)(f(0) + x) = f(x^2) + f(0^2) + 2f(0)$$

Simplifying, we have:

$$f(x)f(0) + xf(x) = f(x^2) + f(0) + 2f(0)$$

$$f(x)f(0) + xf(x) = f(x^2) + 3f(0)$$

### Step 2: Analyzing Special Cases

Assume $f(0) = 0$. The equation simplifies to:

$$xf(x) = f(x^2)$$

Now substitute $y = -x$:

$$(f(x) - x)(f(-x) + x) = f(x^2) + f(x^2) + 2f(-x^2)$$

$$(f(x) - x)(f(-x) + x) = 2f(x^2) + 2f(x^2)$$

Since $xf(x) = f(x^2)$, the above equation becomes $0 = 0$, which does not provide new information, but is consistent.

### Step 3: Full Solution

Substitute $x = y = 0$:

$$f(0)^2 = 3f(0)$$

From this, $f(0) = 0$.

Now, assuming $f(x) = x$ is a candidate solution. Substitute $f(x) = x$ into the original equation:

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

which simplifies directly to:

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

which confirms that:

$$x^2 + 2xy + y^2 = x^2 + y^2 + 2xy$$

Hence, $f(x) = x$ satisfies the equation.

Thus, the only function satisfying the given functional equation is:

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