We prove that the only function satisfying the two conditions is $f(x) = x$.
We prove that $f$ is an injection. If we put $y = 0$ in (i) we get $f(f(x^2) + f(0)) = x^2 + 2f(0)$, for any $x$, or equivalently,
$$f(f(a) + f(0)) = a + 2f(0) \qquad (1)$$
for any number $a \ge 0$. From (1), $f$ is an injection on the set of all nonnegative numbers. We fix now $y$. From the conditions (i) and (ii) we get that $f(\cdot)$ is a superior unbounded function. If $f(y_1) = f(y_2)$ then $f(f(x^2) + y_1 + f(y_1)) = f(f(x^2) + y_2 + f(y_2))$. For large enough values of $x$ the expressions $f(x^2) + y_1 + f(y_1)$ and $f(x^2) + y_2 + f(y_2)$ are positive and thus $y_1 = y_2$.

We prove now that $f(0) = 0$.
Case 1. $f(0) \le 0$. For $a = -2f(0)$, we have $f(f(-2f(a)) + f(0)) = 0$. Thus, there exists a real number $c$ such that $f(c) = 0$.
Plugging $x = 0$ and $y = c$ in (i) we get $f(f(0) + c) = 0$. As $f$ is an injection, we get $f(0) + c = c$, that is $f(0) = 0$.
Case 2. $f(0) \ge 0$. In (i) we put $x = y = 0$: $f(2f(0)) = 2f(0)$. We plug now $a = 3f(0) = f(0) + f(2f(0))$.
From (1) we have $f(a) = f(f(2f(0)) + f(0)) = 2f(0) + 2f(0) = 4f(0)$. We add now $f(0)$ and we get $f(f(a) + f(0)) = f(5f(0)) = 3f(0) + 2f(0) = 5f(0)$.
By plugging now $x = 0$ and $y = 2f(0)$ in (i) we get $f(5f(0)) = 4f(0)$. Thus $f(5f(0)) = 5f(0) = 4f(0)$, from were we obtain $f(0) = 0$.
From (1), for $a \ge 0$ we have $f(f(a)) = a$, and from (i), for $x = 0$, we have $f(y + f(y)) = 2f(y)$ for any real number $y$.
We plug now $y = f(a)$: $f(f(a)+a) = 2f(f(a)) = 2a$ and $y = a$: $f(a+f(a)) = 2f(a)$. We obtained that for any $a \ge 0$ we have $f(a) = a$.
Then, (i) becomes $f(x^2 + y + f(y)) = x^2 + 2f(y)$. We fix now any real $y$. There exists a $x \in \mathbb{R}$ such that $x^2 + y + f(y) > 0$. Then $f(x^2 + y + f(y)) = x^2 + y + f(y) = x^2 + 2f(y)$ and thus $f(y) = y$.