Potla wrote:

Let $f:\mathbb{R}\to \mathbb R$ be a functional equation satisfying
$f(x+y)f(x-y)=\left(f(x)+f(y)\right)^2-4x^2f(y),$
For all $x,y\in\mathbb R,$ where $\mathbb R$ denotes the set of all real numbers.

Let $P(x,y)$ be the assertion $f(x+y)f(x-y)=(f(x)+f(y))^2-4x^2f(y)$

$P(0,0)$ $\implies$ $f(0)^2=4f(0)^2$ $\implies$ $f(0)=0$
$P(x,x)$ $\implies$ $f(x)(f(x)-x^2)=0$ and so $\forall x$ : either $f(x)=0$ , either $f(x)=x^2$

Suppose now that $\exists a\ne 0, b\ne 0$ such that $f(a)=a^2$ and $f(b)=0$
$P(a,b)$ $\implies$ $f(a+b)f(a-b)=a^4\ne 0$
So $f(a+b)\ne 0$ and so $f(a+b)=(a+b)^2$
Same, $f(a-b)\ne 0$ and so $f(a-b)=(a-b)^2$
And our equality becomes $(a^2-b^2)^2=a^4$ and so, since $b\ne 0$ : $b^2=2a^2$
And this is obviously impossible, else :
Choose $c\notin\{0,b,-b,a,-a\}$ :
If $f(c)=0$ , we get (using $a,c$ instead of $a,b$ in lines above : $c^2=2a^2=b^2$ , impossible
If $f(c)=c^2$ , we get, (using $c,b$ instead pf $a,b$ in lines above : $b^2=2c^2$ and so $c^2=a^2$ , impossible.

And so :
either $f(x)=0$ $\forall x$ which indeed is a solution
either $f(x)=x^2$ $\forall x$ which indeed is a soluton.

@SCP : you got " $\forall x$ : either $f(x)=0$ , either $f(x)=x^2$ " and not "either $f(x)=0$ $\forall x$ , either $f(x)=x^2$ $\forall x$ " and you miss the second part of the proof.