By $P(1,1),$ we obtain that $f(0) = 0.$ If $f(2y) = 0,$ for some $y \neq 0,$ then $P(x, y)$ gives that $f(x^2 - y^2) = 0$ for all $x \neq 0.$ As a result, we know that $f(a) = 0$ for all $a > -y^2.$ But then $P(x, a)$ for some huge positive $a$ gives that $f(b) = 0$ for all $b > -a^2,$ and so it's easy to see that we can get that $f \equiv 0$ from this, which works. Otherwise, suppose that $f(y) \neq 0,$ for all $y \neq 0.$ Write $S = \mathbb{R}/\{0\}.$ Let's define the function $g: S \rightarrow S$ by $g(r) = \frac{f(r)}{r},$ for $r \in S.$ Then, the condition rewrites as $g(x^2 - y^2) = g(2x)g(2y),$ for all $x, y \in S$ with $x^2 \neq y^2.$ As a result of the above, whenever we have $a^2 + c^2 = 2b^2$ for $a^2, b^2, c^2$ distinct, we know that $g(2a)g(2b) = g(a^2 - b^2) = g(b^2 - c^2) = g(2b)g(2c),$ so in view of $g(2b) = 0$ we have $g(2a) = g(2c).$ Notice that this works whenever $a, c \in S$ with $a^2 \neq c^2,$ and so this clearly is enough to show that $g$ is constant. Hence, by the definition of $g,$ we know that $f(x) \equiv cx, \forall x \in \mathbb{R},$ and plugging this back shows that $f(x) \equiv 0, x$ are the only two solutions. These two clearly both work, so we're done. $\square$

Vlados021 wrote: Find all functions $f:\mathbb R\to\mathbb R$ such that $$ 4xyf(x^2-y^2)=(x^2-y^2)f(2x)f(2y) $$for all real $x$ and $y$. Let $P(x,y)$ be the assertion $4xyf(x^2-y^2)=(x^2-y^2)f(2x)f(2y)$ $P(1,1)$ $\implies$ $f(0)=0$ If $f(u)=0$ for some $u\ne 0$ : Let $x>0$ : $P(x,\frac u2)$ $\implies$ $f(x^2-\frac{u^2}4)=0$ and so $f(x)=0$ $\forall x\ge 0$ $P(\frac u2,x)$ $\implies$ $f(\frac{u^2}4-x^2)=0$ and so $f(x)=0$ $\forall x\le 0$ And so $\boxed{\text{S1 : }f(x)=0\quad\forall x}$ which indeed is a solution So let us from now consider $f(x)\ne 0$ $\forall x\ne 0$ Let $x\ne 0$ $P(x,\sqrt{x^2+1}-1)$ $\implies$ $f(2x)=2x$ $\forall x\ne 0$, still true when $x=0$ And so $\boxed{\text{S2 : }f(x)=x\quad\forall x}$ which indeed is a solution

Let $P(x,y)$ denote the assertion $4xyf(x^2-y^2)=(x^2-y^2)f(2x)f(2y).$ Clearly $f(0)=0.$ If $f(x_0)=0$ for $x_0\neq 0$ then, $P(x,x_0)$ and $P(x_0,x)$ gives $f\equiv 0.$ So $f(x)=0$ iff $x=0.$ Then $P(\sqrt{x^2+1}+1,x)$ shows $f\equiv \text{Id},$ which fits.

$f(x) \equiv 0$ is a solution. Consider $f(x) \neq 0, \forall x \in \mathbb{R}$ It is obvious that: $f(0) = 0$ and $f(-x) = -f(x) \forall x \in \mathbb{R}$ Let $x \leq 0 \Rightarrow x^2 - 2x \geq 0 $. $P(x, \sqrt{x^2 - 2x}) \Rightarrow 4x.\sqrt{x^2-2x}.f(2x) = 2x.f(2x).f(2\sqrt{x^2-2x}) \Rightarrow 2.\sqrt{x^2-2x}=f(2.\sqrt{x^2-2x}), \forall x \leq 0$ also, $2.\sqrt{x^2-2x}$ receive all value in $\mathbb{R+} \Rightarrow f(x) = x, \forall x \in \mathbb{R+} $. According to $f(x) = -f(-x) $ for all $x \in \mathbb{R}$ and $f(0) = 0 \Rightarrow f(x) = x ,\forall x \in \mathbb{R}$ So, $f(x) = 0 \forall x \in \mathbb{R}$ and $f(x) = x \forall x \in \mathbb{R}$ are solutions Sorry for my worst English. Thanks for reading