Find all functions of defined on the set of all real numbers with real values, such that
$$ f\left(x^2+f(y)\right) = y + f(x)^2 $$
My attempt:
Putting $x = 0$, $f\big(f(y)\big) = y + f^2(0)$ and putting $y = 0$, $f\left(x^2 + f(0)\right) = f(x)^2$.
But I can't go any further.
Suppose $a$ is such that $f(a)=0$ (we know there is such $a$, it is enough to consider $y=-f(x)^2$ for some $x$), then substituting $x=a$ and $y=a$, we have $$f(a^2)=a.$$ Then, set $x=0$ and $y=a^2$, to obtain $$0=f(f(a^2))=a^2+f(0)^2.$$ Since we are only dealing with real numbers, it means that $f(0)=0$, and no other number has $0$ as image.
As consequences, we have $f(f(y))=y$ and $f(y^2)=f(y)^2$ for every $y\in\mathbb{R}$.
This shows $f$ is a bijection. Now plug $y=f^{-1}(\zeta)$, we see $$f(x^2+\zeta)=f^{-1}(\zeta)+(f(x))^2\geq f^{-1}(\zeta)=f(\zeta)$$ Hence $f$ is monotone increasing. Now suppose for some $x_0$ we have $f(x_0)> x_0$, then $x_0=f(f(x_0))\geq f(x_0)> x_0$, a contradiction. Thus we have $f(x)\leq x$. Similarly we have $f(x)\geq x$, so $f(x)=x$ is the only solution.