Beautiful functional equation like always by Dorliri 1.P(0,0) so f(0)=0 2.P(x,-x) so f(z)=z for every negative real number 3.P(0,-x) for x positive real number so f(x^2)=x^2 so f(x)=x for every real number x.

Here is my solution I found during the contest, hopefully I didn't fakesolve it. Edit: I didn't Answer: $f(x)=x \forall x \in \mathbb{R}$ This can be easily verified that it works. Let $P(x,y)$-denote the given assertion. $P(0,0) \implies f(0)=0$ Claim: $f(x)=x \forall x \in \mathbb{R^-}$ Proof: $P(x,-x) \implies f(-x^2)=-x^2$ , if we pick a $y$ s.t $y \in \mathbb{R^+}$ then by taking $x \rightarrow y$ we get: $f(-y^2)=-y^2$, since $y \in \mathbb{R^+}$ we can plug $y \rightarrow \sqrt{y}$ to get: $f(-y)=-y \forall y \in \mathbb{R^+}$, so let $z=-y$ hence $z \in \mathbb{R^-}$ so we are having that $f(z)=z \forall z \in \mathbb{R^-} \implies$ $f(x)=x \forall x \in \mathbb{R^-} \square.$ Claim: $f(x)= x \forall x \in \mathbb{R^+}$ Proof: $P(0,x) \implies f(xf(x))=x^2$, by plugging in $x \rightarrow z$ $\forall z \in \mathbb{R^-}$ we get: $f(zf(z))=z^2$, but from the previous claim we know that $f(z)=z \forall z \in \mathbb{R^-}$ so: $f(z^2)=z^2 \forall z \in \mathbb{R^-}$. Since $z^2 \in \mathbb{R^+}$ we can set $x=z^2$ hence $x\in \mathbb{R^+}$ So $f(x)=x \forall x \in \mathbb{R^+} \square$. Since $f(x)=x \forall x \in \mathbb{R^-}$ and $ f(x)=x \forall x \in \mathbb{R^+}$ we get that $f(x)=x \forall x \in \mathbb{R}$ $\blacksquare$

The only solution is $\boxed{f(x) = x}$, which works. We now show it's the only one. Let $P(x,y)$ be the given assertion. $P(0,0): f(0) = 0$. $P(x,-x): f(x) = f(x) + f(-x^2) + x^2$, so $f(-x^2) = -x^2$, implying that every nonpositive number is a fixed point. For any $y \le 0$, $P(0,y): f(y^2) = y^2$, so every nonnegative number is also a fixed point. Therefore $f$ is the identity.

Let $P(x, y)$ denote the assertion in the problem. $P(x, 0)$ gives $f(x) = f(x) + f(0) \implies f(0) = 0$. $P(x, - x)$ gives $f(x) = f(-x^2) + x^2 \implies f(-x^2) = -x^2 \implies f(a) = a$, when $a$ is non-positive. $P(0, y)$ gives $f(yf(y)) = y^2$. Take y negative, such that $y = -m$ and $m$ is positive. We get; $f(-m) = -m \implies f(m^2) = m^2$, so $f(x) = x$ for all $x$.

Nice classical FE! Standard yet refreshing to see it go the other way, congrats to the author

Put $x=y=0$ to get $f(0)=0$. Then put $x=0$ to get $f(yf(y))=y^2$. Now put $y=-x$ to get $f(-x^2)=-x^2$ hence $f$ is the identity function over negative reals. Take a negative real $k$ and put it in the second equation, then $f(k^2)=k^2$ but $k^2$ is a positive real and takes all positive real values. Hence $f$ is the identity over positive reals as well so $f$ is the identity over all reals.

xoink Let $P(x,y)$ denote the assertion. Consider $P(x,0)$, so $f(0)=0$. Now, by DURR WE WANT TO CANCEL let $x+y=0$ so $f(x+yf(x+y))$ cancels with $f(x)$ and hence $f(xy)=-y^2$ where $x+y=0$, or $f(-y^2)=-y^2$. Hence $f(x)=x$ for all $x\in \mathbb{R}^-$. Now, $P(0,y)$ gives $f(yf(y))=y^2$, if $y$ is negative, then $f(y^2)=y^2$ and $y^2$ spans $\mathbb{R}^+$. Hence $f(x)=x$ for $x\in \mathbb{R}^+$. Combining this with negative reals, and $0$, we conclude $f$ is identity

We claim that the only function is the identity which works. Let $P(x, y)$ denote the assertion. $P(x, 0) \implies f(0) = 0$. $P(x, -x) \implies f(-x^2) = -x^2 \implies f(t) = t \forall t \le 0$. Finally $P(0, x), x < 0 \implies f(yf(y)) = y^2 \implies f(y^2) = y^2 \implies f(x) = x \forall x > $. Therefore $f(x) = x \forall x$ as desired. $\square$