Let $P(x, y)$ denote the usual. By $P\left(x, \frac{x}{2}\right)$ we get $x^2f\left(\frac{3}{4}x^2\right) = \frac{3}{4}x^2f(x)^2$. Through $P\left(-x, \frac{x}{2}\right)$ we see that the right-hand side equals $\frac{3}{4}f(-x)^2$, so for $x \neq 0$ we have $f(-x) = \pm f(x)$. If $f(-x) = f(x)$ for some $x$ then comparing $P(x, y)$ and $P(-x, y)$ we get $4xyf(x^2 - y^2) = 0$, implying $f(z) = 0 \;\forall\; z < x^2$. In this case by setting $y = -1$ we get $-2xf(x^2 - 1) = 0$ implying $f(z) = 0 \;\forall\; z > -1$ and therefore $f(x) = 0$ for all $x$, which is a solution. Now assume that $f(-x) \neq f(x)$, and therefore $f(-x) = -f(x)$, for all nonzero $x$. In particular, $f(x) \neq 0$ for $x \neq 0$. With $P(1, 1)$ we get $2f(0) = 0$, implying that $f(0) = 0$,. Let $y > 0$ and $x = \sqrt{y(y + 2)}$. Then we get $2xyf(2y) = 2yf(x)f(2y)$. Because $2y \neq 0$ implies $f(2y) \neq 0$ we get $f(x) = x$. As $\sqrt{y(y + 2)}$ may take any positive real value this gives $f(x) = x \;\forall\; x > 0$, and because $f(-x) = -f(x)$ this holds for all negative $x$ as well. This gives $f(x) = x \;\forall\; x$ which is clearly a solution.

mathisreal wrote: Find all functions $f:\mathbb R \rightarrow \mathbb R$, such that $2xyf(x^2-y^2)=(x^2-y^2)f(x)f(2y)$ $$2xyf(x^2-y^2)=(x^2-y^2)f(x)f(2y)...(\alpha)$$In $(\alpha) x=y, x\neq 0:$ $$\Rightarrow 2x^2f(0)=0$$$$\Rightarrow f(0)=0$$In $(\alpha) y=2x, x\neq 0:$ $$\Rightarrow 4x^2f(3x^2)=3x^2f(2x)^2$$$$\Rightarrow 4f(3x^2)=3f(2x)^2...(\beta)$$In $(\alpha) y=-2x, x\neq 0:$ $$\Rightarrow -4x^2f(3x^2)=3x^2f(-2x)f(2x)$$By $(\beta):$ $$\Rightarrow f(2x)^2=f(-2x)f(2x)$$$$\Rightarrow f(2x)=0 \text{ or } f(-2x)=-f(2x)...(\lambda)$$If $\exists a\neq 0 / f(a)=0, a\neq 0:$ In $(\alpha) x=a, y\neq 0':$ $$\Rightarrow f(a^2-y^2)=0$$$$\Rightarrow f(x)=0, \forall x<a^2...(\theta)$$In $(\alpha) y=\frac{a}{2}, x\neq 0:$ $$\Rightarrow f(x^2-\frac{a^2}{4})=0$$$$\Rightarrow f(x)=0, \forall x>-\frac{a^2}{4}...(\omega)$$By $(\theta)$ and $(\omega):$ $$\Rightarrow f(x)=0, \forall x\in\mathbb{R}$$Else: By $(\lambda):$ $$\Rightarrow -f(2x)=f(-2x)$$$$\Rightarrow f\text{ is odd}$$For a fixed $y\ge-2$ we make $x$ such that $x^2-y^2=2y$ $$\Rightarrow x^2=y(y+2)$$In $(\alpha):$ $$\Rightarrow f(x)=x, \forall x\ge0$$In $(\alpha) |x|\ge|y|, y\ge 0:$ $$\Rightarrow f(x)=x, \forall x\in\mathbb{R}$$ $$\Rightarrow f(x)=0 \text{ and }f(x)=x \text{ are the only solutions}_\blacksquare$$