Obviously $f$ is surjective and $1-1$ Let $f(t)=0$ then $P(t,t),P(0,t^2)$ gives $t=0$ $P(0,y)$ gives $f(f(y))=y$ $P(x,0)$ gives $f(x^2)=xf(x)$ $P(f(x),y)$ and $P(x,y)$ using $1-1$ gives $f(x)=+-x$ Supose that there is $a,b$9diferen from $0$) such that $f(a)=a$ and $f(b)=-b$ then $P(a,b)$ contradiction So $f(x)=x$ for all $x$ or $f(x)=-x$ for all $x$

Let $P(x,y)$ denote the assertation of $(x,y)$ into the functional equation. Notice that $P(0,y)\Longrightarrow f(f(y))=y\text{ }\forall y\in\mathbb{R}\quad (1)\Longrightarrow f\text{ is bijective}$. Now we have: \[P(f(0),f(0)):\quad f\left(f(0)^2+f(f(0))\right)=f(0)+f(0)f(f(0))\Longrightarrow f(f(0)^2)=f(0)\Longrightarrow f(0)^2=f(f(0))=0\Longrightarrow f(0)=0\]\[P(x,0):\quad f(x^2)=xf(x)\quad (2)\]\[\Longrightarrow x^2\overset{(1)}{=}f(f(x^2))\overset{(2)}{=}f(xf(x))\overset{(1)}{=}f(f(f(x))f(x))=f(f(x)f(f(x)))\overset{(2)}{=}f(f(f(x)^2))\overset{(1)}{=}f(x)^2\Longrightarrow f(x)=\pm x \text{ }\forall x\in\mathbb{R}\]Assume that there exist $u\neq 0,v\neq 0$, $u,v \in\mathbb{R}$ such that $f(u)=u$, $f(v)=-v$. Then we have a contradiction by: \[P(u,v):\quad f(u^2-v)=f(u^2+f(v))=v+uf(u)=v+u^2\Longrightarrow u^2+v=\pm (u^2-v)\Longrightarrow u=0\text{ or } v=0\]Therefore $f(x)=x$ $\forall x\in\mathbb{R}$ or $f(x)=-x$ $\forall x\in\mathbb{R}$ are the only possible solutions. These are indeed solutions because if $f(x)=x$ $\forall x\in\mathbb{R}$, then $f(x^2+f(y))=x^2+y=y+xf(x)$ and if $f(x)=-x$ $\forall x\in\mathbb{R}$, then $f(x^2+f(y))=-x^2+y=y+xf(x)$, so the only solutions are $\boxed{f(x)=x\text{ }\forall x\in\mathbb{R}}$ and $\boxed{f(x)=-x\text{ }\forall x\in\mathbb{R}}$.

Define $P(x,y)$ as usual. $P(0,y) \implies f(f(y))=y \implies f$ is bijective. $P(f(x),y) \implies f(f(x)^2+f(y)) = y+f(x)f(f(x)) = y+xf(x) = f(x^2+f(y))$ Since $f$ is injective, $f(f(x)^2+f(y)) = f(x^2+f(y)) \implies f(x)^2 + f(y) = x^2 + f(y)$ So, $f(x)^2 = x^2$ and $f(x) = \{x, -x\}$ BEWARE OF THE POINTWISE TRAP!!! Now, suppose $\exists$ a, b such that $f(a) \neq a$ and $f(b) \neq -b$. This implies that $f(a) = -a$ and $f(b) = b$ $P(a,b) \implies f(a^2+f(b)) = b+af(a)$ This simplifies to $f(a^2+b) = b-a^2$ Now, we have $2$ cases. Case 1: $f(a^2+b)=a^2+b$ The equation becomes $a^2+b=b-a^2$ and this gives $a=0$, which contradicts the fact that $f(a) \neq a$ (since $f(0) =0$). Case 2:$f(a^2+b)=-a^2-b$ The equation becomes $-a^2-b=b-a^2$ and this gives $b=0$, which contradicts the fact that $f(b) \neq -b$ (since $f(0) =0$). Thus, we cannot have a mixed case and hence conclude that either $\boxed{f(x)=x \, \forall x \in \mathbb{R}}$ or $\boxed{f(x)=-x \, \forall x \in \mathbb{R}}$ The two solutions can be easily verified.

Either $f(x)=x$ for all $x$; or $f(x)=-x$ for all $x$. Denote the assertion by $P(x,y)$. $P(0,x)$ yields $f(f(y))=y$. In particular, $f$ is injective. Now, $P(f(x),y)$ yields $f\bigl(f(x)^2+f(y)\bigr)=y+xf(x)$. Comparing this with $P(x,y)$ and recalling injectivity of $f$, we get $f(x)^2=x^2$ for all $x$. Clearly the aforementioned functions work; let us prove there are no `intermediate' solutions. To that end, assume there is an $f$ with $f(x_0)=x_0$, $f(y_0)=-y_0$ and $x_0y_0\ne 0$. Considering $P(x_0,y_0)$ we find \[ f\bigl(x_0^2 - y_0\bigr) = x_0^2+y_0. \]Now that $f(x_0^2-y_0)\in\{x_0^2-y_0,-x_0^2+y_0\}$, both of these contradict with $x_0y_0\ne 0$.

$P(0,x)$ gives $f(f(x))=x$ so $f$ is bijective. $P(f(x),y)$ gives $f(f(x)^2+f(y))=y+f(x)f(f(x))=y+xf(x)=f(x^2+f(y))$ Injectivity gives $f(x)^2=x^2$. There's a pointwise trap. Let $f(a)=a, f(b)=-b$, $P(a,b)$ gives $f(a^2-b)=a^2+b$ which is a contradiction. So the only solutions are $f(x)=x$ and $f(x)=-x$.

The only solutions are $\boxed{f(x) = x}$ and $\boxed{f(x) = -x}$. These work. Now we prove they are the only ones. Let $P(x,y)$ be the given assertion. $P(0,x): f(f(x)) = x$, so $f$ is bijective. $P(f(x), y): f(f(x)^2 + f(y)) = f(x^2 + f(y))$, so $f(x)^2 = x^2$, which means $f(x)\in \{-x,x\}$. If $f(a) = a$ and $f(b) = -b $ for some $a,b\ne 0$ then \[P(a,b): f(a^2 - b) = a^2 + b,\], which implies $a = 0$ or $b = 0$, contradiction. Thus, $f(x) = -x$ for all $x$ or $f(x) = x$ for all $x$.

Let $P(x,y)$ denote the assertation $f(x^2 +f(y))=y+xf(x)$ $P(0,y)$ $\Rightarrow$ $f(f(y))=y$ $\Rightarrow$ $f$ is bijective Let $f(t)=0,f(0)=t$ $\Rightarrow$ $P(t,t)$ $\Rightarrow$ $f(t^2)=t=f(0)$ $\Rightarrow$ $t^2=0$ $\Rightarrow$ $f(0)=0$ $P(x,0)$ $\Rightarrow$ $f(x^2)=xf(x)$ $P(x,f(y))$ $\Rightarrow$ $f(x^2+y)=f(y)+f(x^2) \text{ }\forall x,y\in\mathbb{R}$ $f((x+1)^2)=(x+1)f(x+1)=(x+1)(f(x)+f(1))=xf(x)+f(x)+xf(1)+f(1)$ $f((x+1)^2)=f(x^2+2x+1)=f(x^2)+f(2x)+f(1)=f(x^2)+2f(x)+f(1)$ From the above two we get: $f(x)=xf(1)\text{ }\forall x\in\mathbb{R}$ $P(f(1),0)$ $\Rightarrow$ $f(f(1)^2)=f(1)$ $\Rightarrow$ $f(1)^2=1$ $\Rightarrow$ $f(1)=\pm 1$ After checking we get: $\boxed{f(x)=x\text{ }\forall x\in\mathbb{R}}$ and $\boxed{f(x)=-x\text{ }\forall x\in\mathbb{R}}$