Let $f : \mathbb{R} \to \mathbb{R}$ be a function such that for all $x, y \in \mathbb{R}$, the following functional equation holds:
$$f(x^2 + y^2 + 2f(xy)) = (f(x+y))^2.$$

We need to find all possible functions $f$ that satisfy this equation.

### Step 1: Consider simple test cases

First, set $x = y = 0$:

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

Let $f(0) = c$. Then, we have:
$$f(2c) = c^2.$$

### Step 2: Analyze specific function candidates

#### Case 1: Assume $f(x) = 0$

Substitute $f(x) = 0$ for all $x$:
$$f(x^2 + y^2 + 2 \cdot 0) = 0 = (0)^2.$$
This satisfies the functional equation.

#### Case 2: Assume $f(x) = x$

Substitute $f(x) = x$:
$$f(x^2 + y^2 + 2xy) = f((x+y)^2) = (x+y)^2.$$
This implies
$$x^2 + y^2 + 2xy = (x+y)^2,$$
which is true generally. Therefore, $f(x) = x$ is another solution.

### Step 3: Consider functions of binary nature

For $f$ of the form:
$$f(x) = \begin{cases} 1, & x \notin X, \\ -1, & x \in X, \end{cases}$$
where $X \subset (-\infty, -\frac{2}{3})$.

- When $x, y \notin X$:
$$f(x^2 + y^2 + 2f(xy)) = f(x^2 + y^2 + 2 \cdot 1) = 1 = 1^2.$$
- When $x, y \in X$:
$$f(x^2 + y^2 + 2f(xy)) = f(x^2 + y^2 + 2 \cdot (-1)) = 1 = (-1)^2,$$
given the structure of $X$.

The functions $f(x) = x$, $f(x) = 0$, and the binary functions described match all conditions provided by the problem statement.

Thus, the set of all functions $f$ satisfying the given condition is:
$$f(x) = x, \quad f(x) = 0, \quad \text{and functions of the form} \quad f(x) = \begin{cases} 1, & x \notin X, \\ -1, & x \in X, \end{cases} \text{ where } X \subset (-\infty, -\frac{2}{3}).$$

Therefore, the set of all solutions is:
$$\boxed{\{ f(x) = x, f(x) = 0, f(x) = \begin{cases} 1, & x \notin X, \\ -1, & x \in X, \end{cases} \text{ where } X \subset (-\infty, -\frac{2}{3}) \}}$$