f(x² + f(y)) = y + f(x)²

Let $a=f(0)$.

Step 1. $x=0$: $f(f(y))=y+a^2$. Thus $f$ is injective and surjective (the right side takes every real value).

Step 2. By surjectivity pick $t$ with $f(t)=0$. Putting $y=t$ gives
$$f(x^2)=t+f(x)^2\qquad(\ast)$$
for all $x$.

Step 3: oddness. Replacing $x$ by $-x$ in $(\ast)$ gives $f(x)^2=f(-x)^2$, so $f(-x)=\pm f(x)$. If $x\neq0$, the case $f(-x)=f(x)$ contradicts injectivity. Hence $f(-x)=-f(x)$ for all $x\ne0$. If $t\ne0$, then $f(-t)=-f(t)=0=f(t)$, contradicting injectivity again. Therefore $t=0$, i.e. $f(0)=0$ and $f$ is odd.

Step 4. Now $a=0$, so $f(f(y))=y$ and $(\ast)$ becomes $f(x^2)=f(x)^2\ge0$: $f\ge0$ on $[0,\infty)$. The original equation reads $f(x^2+f(y))=y+f(x^2)$. For every $u\ge0$ (taking $x=\sqrt u$): $f(u+f(y))=y+f(u)$; replacing $y$ by $f(y)$,
$$f(u+y)=f(u)+f(y)\qquad(u\ge0,\ y\in\mathbb{R}).$$
For $u<0$, use oddness: $f(u+y)=-f((-u)+(-y))=-f(-u)-f(-y)=f(u)+f(y)$. So $f$ is additive on $\mathbb{R}$.

Step 5. An additive function that is $\ge0$ on $[0,\infty)$ is non-decreasing, hence $f(x)=cx$ (Cauchy with monotonicity). From $f(f(x))=x$: $c^2=1$, and $f(1)=c\ge0$ gives $c=1$.

Answer. $f(x)=x$ (check: $f(x^2+y)=x^2+y$ ✓).

Lesson. A strong sequence of moves: injective $\to$ locate the zero $\to$ oddness $\to$ $f(f(y))=y$ $\to$ additivity $\to$ monotonicity from nonnegativity of squares.