All solutions are $f(x)=0,$ $f(x)=\frac{1}{4}x^2,$ and $f(x)=-\frac{1}{4}x^2.$ Denote the given equation as $P(x, y).$ $\boxed{f(x)=0}$ is one solution. $P(0, 0)$ yields $f(0)=0.$ Suppose there exist $c\neq 0$ such that $f(c) \neq 0.$ Claim. $f$ is even. Proof. From $P(c, y)$ we have that $f(\mathbb{R})$ contains all reals whose sign is same as $f(c).$ Hence $f(x+y) - f(x-y)$ can represent all positive reals from the fact that $(x+y, x-y)$ can produce any ordered pair of $\mathbb{R}^2.$ Now compare $P(x, y)$ and $P(x, -y)$, we have \[ f(f(x+y)-f(x-y)) = y^2f(x) = f(f(x-y)-f(x+y)). \]Since $f(x+y)-f(x-y)$ spans all reals, claim is proved. To finish, compare $P(x, c)$ and $P(c, x)$ then we have \[ c^2f(x) = f(f(x+c) - f(x-c)) = f(f(c+x) - f(c-x)) = x^2f(c), \]so $\forall\;x\in \mathbb{R},\;\; f(x) = kx^2$ for some $k.$ Plugging this into original equation yields $k\in \{\frac{1}{4}, -\frac{1}{4}\},$ which result in two more solutions, $\boxed{f(x) = \frac{1}{4}x^2}$ and $\boxed{f(x) = -\frac{1}{4}x^2}$

mjk20016 wrote: Comparing $P(x, x-t)$ and $P\left(x-\frac{t}{2}, x-\frac{t}{2}\right)$ we have \[ (x-t)^2f(x) = f(f(2x-t)) = \left(x-\frac{t}{2}\right)^2f(x). \]How you got it? I think there should be \[ (x-t)^2f(x) = f(f(2x-t)) = \left(x-\frac{t}{2}\right)^2f(x-\frac{t}{2}). \]

Here's my approach. The only solutions are $\boxed{f(x)=0, f(x)=\frac{x^2}{4}, f(x)=-\frac{x^2}{4}}$, which clearly work. Let $P(x, y)$ be the assertion to the given equation. Note that $f(x)=0$ satisfies the given condition. Now, suppose that there exists $a \in \mathbb{R}$ such that $f(a) \ne 0$. $P(x, 0)$ gives $f(0)=0$. $P(a, y)$ gives $$f(f(a+y)-f(a-y))=y^2f(a)$$for all $y \in \mathbb{R}$. Hence, the function $f$ can take either any positive reals (if $f(a)>0$) or any negative reals (if $f(a)<0$). By $P(x, x)$ and $P(x, -x)$, we have $$f(f(2x))=x^2f(x)=f(-f(2x)),$$and since $f(2x)$ can take either any positive reals or any negative reals, $f(x)=f(-x)$ for all $x \in \mathbb{R}$. Observe that $P(y, x)$ implies \begin{align*}x^2f(y)&=f(f(x+y)-f(y-x)) \\&=f(f(x+y)-f(x-y))=y^2f(x).\end{align*}Therefore, $$\frac{f(x)}{x^2}=\frac{f(y)}{y^2}$$holds for any real numbers with $xy \ne 0$, which gives $f(x)=cx^2$ for all nonzero real numbers. Since $f(0)=0$, we have $f(x)=cx^2$ for any real numbers and putting $f(x)=cx^2$ in the given equation, we get $c=0, \pm \frac{1}{4}$. Hence, we're done.

Olympiadium wrote: $P(a, \frac{y}{f(a)})$ gives $$f(f(a+\frac{y}{f(a)})-f(a-\frac{y}{f(a)}))=\boxed{y^2}$$for all $y \in \mathbb{R}$. Shouldn't there be $\boxed{\frac{y^2}{f(a)}}$ instead of $\boxed{y^2}$ ? Hence, the function $f$ can take any nonnegative reals. $\boxed{How\ did\ you\ get\ that?}$

Let $f(j)\ne0$ for some $j$ (the only alternative is $\boxed{f(x)=0}$ which is a solution) and $P(x,y)$ denote the given assertion. Note that $f$ fits the equation iff $-f$ does by negating $y$ in the assertion, so WLOG $f(j)>0$. If $x>0$, we have: $P\left(j,\sqrt{\frac x{f(j)}}\right)\Rightarrow f\left(f\left(j+\sqrt{\frac x{f(j)}}\right)-f\left(j-\sqrt{\frac x{f(j)}}\right)\right)=x$ so $f$ is surjective over $\mathbb R^+$. $P(0,0)\Rightarrow f(0)=0$ $P\left(\frac x2,\frac x2\right)-P\left(\frac x2,-\frac x2\right)\Rightarrow f(f(x))=f(-f(x))$ Since $f$ is surjective, $f$ must be even. $P(y,x)-P(x,y)\Rightarrow x^2f(y)=y^2f(x)\Rightarrow\frac{f(x)}{x^2}=c\Rightarrow f(x)=cx^2$, testing gives the only new solutions $\boxed{f(x)=\frac14x^2}$ and $\boxed{f(x)=-\frac14x^2}$.

mjk20016 wrote: All solutions are $f(x)=0,$ $f(x)=\frac{1}{4}x^2,$ and $f(x)=-\frac{1}{4}x^2.$ Denote the given equation as $P(x, y).$ $\boxed{f(x)=0}$ is one solution. $P(0, 0)$ yields $f(0)=0.$ Suppose there exist $c\neq 0$ such that $f(c) \neq 0.$ Claim. $f$ is even. Proof. From $P(c, y)$ we have that $f(\mathbb{R})$ contains all reals whose sign is same as $f(c).$ Hence $f(x+y) - f(x-y)$ can represent all positive reals from the fact that $(x+y, x-y)$ can produce any ordered pair of $\mathbb{R}^2.$ Now compare $P(x, y)$ and $P(x, -y)$, we have \[ f(f(x+y)-f(x-y)) = y^2f(x) = f(f(x-y)-f(x+y)). \]Since $f(x+y)-f(x-y)$ spans all reals, claim is proved. To finish, compare $P(x, c)$ and $P(c, x)$ then we have \[ c^2f(x) = f(f(x+c) - f(x-c)) = f(f(c+x) - f(c-x)) = x^2f(c), \]so $\forall\;x\in \mathbb{R},\;\; f(x) = kx^2$ for some $k.$ Plugging this into original equation yields $k\in \{\frac{1}{4}, -\frac{1}{4}\},$ which result in two more solutions, $\boxed{f(x) = \frac{1}{4}x^2}$ and $\boxed{f(x) = -\frac{1}{4}x^2}$ I have a question, but how did you get that f(x+y)-f(x-y) spands for all reals?

The only solutions are $f\equiv 0$, $f(x) = \frac 14 x^2$, and $f(x) = - \frac 14 x^2$, which work. Now we show they are the only solutions. Clearly $f$ constant works. Now assume $f$ isn't constant. $P(0,0): f(0) = 0$. Claim: $|f|$ is surjective over the nonnegative reals. Proof: Fix $x$ with $f(x) \ne 0$. Note that $y^2 f(x)$ is surjective over all real numbers either $ \ge 0$ or $\le 0$, as desired. $\square$ Claim: $f$ is even. Proof: $P(x,x)$ compared with $P(x,-x)$ gives that $f(f(2x)) = f(-f(2x))$. Now fix any nonnegatve real $r$ and $x$ with $|f(2x)| = r$. Note that $f(r) = f(-r)$, so $f$ is even, as desired. $\square$ $P(x,y)$ compared with $P(y,x)$ gives that $x^2 f(y) = y^2 f(x)$, so $\frac{f(x)}{x^2} = \frac{f(y)}{y^2}$ for all $x,y \ne 0$. Thus, $f(x) = cx^2$ for some constant $c$ and all reals $x$ (since it's obviously true for $x = 0$ as we had $f(0)= 0$). We now have $f(c((x+y)^2 - (x-y)^2) ) = c x^2 y^2$, so $f(4cxy) = c x^2y^2$, meaning that $16c^3 = c \implies c^2 = \frac{1}{16}$, so $c \in \left\{- \frac 14, \frac 14 \right \}$, as desired.