$\clubsuit \color{magenta}{\textit{\textbf{ANS:}}}$ $f(x)=0 \quad \textrm{and} \quad f(x)=x \quad \forall x\in \mathbb{R}$ .
$\blacklozenge \color{blue}{\textit{\textbf{Proof:}}}$ It's easy to see that these are indeed solutions to the given FE. Let $P(x,y)$ be the given assertion, we have $P(x,0): f(x)(f(-1)+1)=x^2f(0)$ if $f(-1)\ne -1$ , we will have $f(x)=cx^2$ for some real number $c$ , plugging it into our FE, $c^4x^6y^2+\ldots =c^2x^2(cx^2y-1)^2=cx^2y^2-cx^2$ which means $c=0$ and $f(x)=0$ for all real $x$ . If $f(-1)=-1$ , then $f(0)=0$ . If there exists a real number $u$ , such that $f(u)=0$ , then $P(u,x): u^2f(x)=0 \quad \forall x\in \mathbb{R}$ which if $f(x)\ne 0$ for all real $x$ , then $u=0$ . So, $f$ is injective at $0$ and $f$ is not the zero function. Then, $P(x,x): f(xf(x)-1)=x^2-1 \quad \forall x\ne 0$ but since $f(0)=0$ , $f(xf(x)-1)=x^2-1 \quad \forall x\in \mathbb{R}$ . Now, by comparing $P(x, 1)$ and $P(1,f(x))$ , we have $\frac{x^2-f(x)}{f(x)}=f(f(x))-1 \implies f(x)f(f(x))=x^2 \quad \forall x\ne 0$ but again $f(0)=0$ , so this also holds for all real $x$ . Plugging this back to our original FE, we have $P(x,y):f(yf(x)-1)=f(f(x))f(y)-1$ and $P(1,x+1): f(x+1)=f(x)+1.$ Most importantly, $f$ is odd since $P(-1,x): f(-x)-1=f(-x-1)=-f(x)-1 \implies f(-x)=-f(x) \quad \forall x\in \mathbb{R}.$ Therefore, since $f$ is odd and $f(xf(x)-1)=x^2-1$ , $f$ is surjective over $\mathbb{R}$ . This means $P(x,f(y)): f(f(x)f(y))=f(f(x))f((y)) \implies f(xy)=f(x)f(y)$ $f$ is multiplicative and in particular, $f(x^2)=f(x)^2$ . Finally, as $\begin{cases} f(x+1)=f(x)+1 \\ f(x^2)=f(x)^2, \end{cases}$ this implies $f(x)=x$ (it's a well-known FE). $\quad \blacksquare$