*Answer:* $f(x) = 0$ and $f(x) = x$.

Substituting $x = y = 0$ to the equation, we get $(f(0))^2 = 0 + 2f(0) + (f(0))^2$. Simplifying this gives $f(0) = 0$.

Substituting $y = 0$ to the original equation, we get the equation $(f(x))^2 = x f(x) + 2 f(0) + (f(0))^2$ which must be satisfied for all real $x$. Since $f(0) = 0$, this simplifies to $f(x)(f(x) - x) = 0$. Thus, for each $x$, either $f(x) = 0$ or $f(x) = x$.

Assume there exists a real number $c \neq 0$ such that $f(c) = 0$. Substituting $x = c$ to the original equation, we get $(f(c+y))^2 = 2f(cy) + (f(y))^2$, or
$$2f(cy) = (f(c + y))^2 - (f(y))^2. \quad (1)$$
This is valid for any $y$. Let's analyse four cases based on whether $f(c+y) = 0$ or $f(c+y) = c+y$ and whether $f(y) = 0$ or $f(y) = y$.

1) If $f(c+y) = 0$ and $f(y) = 0$, then (1) simplifies to $f(cy) = 0$.

2) If $f(c+y) = 0$ and $f(y) = y$, then (1) gives $f(cy) = -\frac{y^2}{2}$. Assuming that $f(cy) = cy \neq 0$, we get $c = -\frac{y}{2}$ i.e. $y = -2c$.

3) If $f(c+y) = c+y$ and $f(y) = 0$, then (1) gives us $2f(cy) = (c+y)^2$. Assuming that $f(cy) = cy$, we get $c^2 + y^2 = 0$ from which $c=0$, contradiction.

4) If $f(c+y) = c+y$ and $f(y) = y$, then (1) gives us $2f(cy) = c^2 + 2cy$. Assuming that $f(cy) = cy$, we get $c^2 = 0$, contradiction.

Thus $f(cy) \neq 0$ can only be valid if $y = -2c$, i.e. $f(x) \neq 0$ can only be valid if $x = c \cdot (-2c) = -2c^2$. Thus it is possible to choose a real number $d$ such that $d \neq 0$, $f(d) = 0$, and $|d| \neq |c|$. By replacing $d$ by $c$ we can analogously conclude that $f(x) \neq 0$ can only be valid when $x = -2d^2$. As $-2d^2 \neq -2c^2$, $f(x) \neq 0$ cannot be valid for any $x \neq 0$. So $f(x) = 0$ for all $x$.

Therefore the only suitable functions are $f(x) = 0$ and $f(x) = x$.