The only functions which work are the identity and the zero-function. It is clear that both satisfy the equation. Now we show any valid function that is not zero is the identity. Suppose $f(x_0) \ne 0$ for some $x_0 \in \mathbb{R}$. If $f(0) \ne 0$ then $f(0)f(yf(0)-1)=-f(0)$ but $f \equiv -1$ is not a solution. So $f(0)=0$. Put $x=x_0, y=0$ so $f(-1)=-1$. Put $y=x$ so $f(xf(x)-1)=x^2-1$ for all $x$; hence $f$ is surjective over $(-1, \infty)$. Suppose $f(a)=0$ then $x=a, y=x_0$ implies $a=0$. So $f$ is injective at $0$. Now $f(f(1)-1)=0$ so $f(1)=1$. Put $x=1$ to get $f(y-1)=f(y)-1$ for all $y$ so $f(y-N)=f(y)-N$ for all integers $N \ge 0$. Now $f(y)=f(y+N)-N$ so $f(y+N)=f(y)+N$ for all integers $N \ge 0$. Consequently, $f$ is surjective over all of $\mathbb{R}$. Plug $f(yf(x)-1)=f(yf(x))-1$ to conclude $f(x)f(yf(x))=x^2f(y)$. Put $y=1$ so $x^2=f(x)f(f(x))$ hence $f(yf(x))=f(y)f(f(x))$ for all $x \ne 0$. For $x=0$, last claim is obvious. Now surjectivity shows $f(ty)=f(t)f(y)$ for all $t,y \in \mathbb{R}$ so $f$ is multiplicative. Thus, $f(z^2)=f(z)^2 \ge 0$ and $f(-z^2)=f(-1)f(z)^2=-f(z)^2 \le 0$ hence $f$ preserves the sign of the argument. Now pick $x>1$ and so $1 \ge x-\lfloor x \rfloor \ge 0$ hence $f(x-\lfloor x \rfloor) \ge 0$ and $f(x-\lfloor x \rfloor-1) \le 0$ hence $$\lfloor x \rfloor \le f(x) \le \lfloor x \rfloor+1$$so $$x-1 \le f(x) \le x+1$$for all $x>1$. Now suppose $f(t_0) \ne t_0$ for some $t_0>1$. 1. If $f(t_0)>t_0$. Now for $y>1$ sufficiently large, $$yf(t_0)^2-2f(t_0) \le f(t_0)f(yf(t_0)-1)=t_0^2f(y)-f(t_0) \le yt_0^2+t_0^2-f(t_0)$$hence $$y \le \frac{t_0^2+f(t_0)}{f(t_0)^2-t_0^2}$$which fails as $y \rightarrow \infty$. 2. If $f(t_0)<t_0$. Again for $y>1$ sufficiently large, $$yf(t_0)^2 \ge f(t_0)f(yf(t_0)-1)=t_0^2f(y)-f(t_0) \ge yt_0^2-t_0^2-f(t_0)$$hence $$y \le \frac{t_0^2+f(t_0)}{-f(t_0)^2+t_0^2}$$which fails as $y \rightarrow \infty$. Finally, shifting down by large integers $N$, we obtain $f(x)=x$ for all $x \in \mathbb{R}$ as desired. $\blacksquare$

Easy for P3 (if problems order in difficult)and simple solution. Let $P(x,y)$ be the assertion of $f(x)f(yf(x)-1)=x^2f(y)-f(x)$ $P(0,0)\to f(0)\cdot f(-1)=-f(0),$ then $1) f(0)=0,$ or $f(-1)=-1.$ $1)$ $P(x,0)\to f(x)\cdot f(-1)=-f(x).$ $1.1)$ For all $x,$ $f(x)\equiv 0,$ Indeed this solution work. $1.2)$ $\exists a\in\mathbb{R} ,$ such that $f(a)\not= 0,$ them from $P(a,0)\to f(-1)=-1.$ $2)$ $P(-1,0)\to f(0)=0.$ Then from both condition we can get $f(0)=0,f(-1)=-1.$ Lemma: $f(a)\equiv 0 \iff a\equiv 0.$ Proof: we know $f(0)=0,$ let show $f(a)=0\to a=0.$ $P(a,-1)\to -a^2=0\to a=0.$ As desired. From $P(1,1)\to f(1)\cdot f(f(1)-1)=0.$ $2.1)$ $f(1)=0.$ From $P(1,x)\to $ for all $x,$ $ f(x)=0,$ but $f(-1)=-1.$ contradiction. $2.2)$ $f(1)\not= 0\to f(f(1)-1)=0.$ From lemma we get $f(1)=1.$ Then $P(1,x+1)\to f(x+1)=f(x)+1,$ (or $f(x-1)=f(x)-1.$) Then our equation equivalent to $Q(x,y):f(x)\cdot f(yf(x))=x^2f(y).$ From $Q(x,1)\to f(x)\cdot f(f(x))=x^2.$ Also from $Q(x+1,1)\to (f(x)+1)\cdot f(f(f(x))+1)=f(x)\cdot f(f(x))+f(x)+f(f(x))+1=x^2+2x+1,$ then we get $f(f(x))=2x-f(x).$ From $f(x)\cdot f(f(x))=2x,$ use $f(f(x))=2x-f(x),$ we get $(x-f(x))^2=0.$ Then $f(x)\equiv x,$ for all $x.$ Indeed this solution work.

Nice problem. Here's my solution: Let $P(x,y)$ denote the given assertion. Note that the only constant function which satisfies the given equation is the zero function. So from now on we assume that $f$ is non-constant. CLAIM For $c \in \mathbb{R}$, we have $f(c)=0 \Leftrightarrow c=0$. Proof: Note that we have $$P(-1,0) \Rightarrow f(-1)^2=f(0)-f(-1) \Rightarrow f(-1)(f(-1)+1)=f(0)$$Also, $$P(0,0) \Rightarrow f(0)f(-1)=-f(0) \Rightarrow f(0)=0 \text{ OR } f(-1)=-1$$If $f(0)=0$, then using the first relation, we get that either $f(-1)=0$ or $f(-1)=-1$. But, when $f(-1)=0$, we have $P(x,0) \Rightarrow f(x)=0$, which contradicts the fact that $f$ is non-constant. That means $f(0)=0$ implies $f(-1)=-1$. And, when $f(-1)=-1$, then (again from the first relation), we get $f(0)=0$. Summarizing the above, we can say that $f(0)=0$ and $f(-1)=-1$ are both true simultaneously. Now, suppose that $f(c)=0$. Then $P(c,-1) \Rightarrow c^2f(-1)=0 \Rightarrow c=0$. $\Box$ Return to the problem at hand. Then we get $P(x,-1) \Rightarrow f(x) \cdot f(-f(x)-1)=-x^2-f(x)$ $$P(-1,f(x)) \Rightarrow -f(-f(x)-1)=f(f(x))+1 \Rightarrow f(x) \cdot f(-f(x)-1)=-f(x) \cdot (f(f(x))+1)$$$$\Rightarrow x^2+f(x)=f(x) \cdot (f(f(x))+1) \Rightarrow f(f(x)) \cdot f(x)=x^2 \text{ } (*)$$ Now, $P(1,1) \Rightarrow f(1) \cdot f(f(1)-1)=0$. Using our Claim, we get that $f(1)-1=0 \Rightarrow f(1)=1$ $($as $f(1) \neq 0)$. Then $$P(1,y) \Rightarrow f(y-1)=f(y)-1 \Rightarrow \text{ By an easy induction, }f(y+n)=f(y)+n \text{ } \forall n \in \mathbb{N}$$ Finally, Putting $x \Longrightarrow x+n$ in $(*)$, and using the above relation, we get that $$(f(f(x))+n)(f(x)+n)=x^2+2nx+n^2=f(f(x)) \cdot f(x)+2nx+n^2$$$$\Rightarrow n(f(f(x))+f(x))=2nx \Rightarrow f(f(x))+f(x)=2x \text{ } (**)$$ Using $(*)$ and $(**)$, we can easily find that $f(f(x))=f(x)=x$. Thus, our final answers are $$\boxed{f \equiv 0 \text{ AND } f(x)=x \text{ } \forall x \in \mathbb{R}}$$Now one can easily verify that these solutions actually work. Hence, done. $\blacksquare$

I think this solution is more elementary than the others, although many parts are similar to the given above.

Sorry for double posting, but I have a different ending after proving the claim in post #4. Here's my solution: As proved in post #4, we have $f(0)=0,f(-1)=-1$ and that $f$ is non-constant. Now, $$P(-1,y) \Rightarrow -f(-y-1)=f(y)+1 \overset{y \rightarrow -y}{\Longrightarrow} f(y-1)=-f(-y)-1$$Again, as shown in post #4, one easily gets that $f(y-1)=f(y)-1$. Thus, we have $f(-y)=-f(y)$, i.e. $f$ is odd. We will show that $f$ is injective also. Suppose we have $f(a)=f(b)$ for some $a,b \in \mathbb{R}$. Then $$P(a,-1)-P(b,-1) \Rightarrow a^2=b^2 \Rightarrow b=a \text{ OR } b=-a$$However, we cannot have $f(-a)=f(a)$, as $f$ is odd. Hence, $f$ must be injective. Now, as $f(y-1)=f(y)-1$, so we can rewrite the problem condition as $$P(x,y):= f(x)(f(yf(x))-1)=x^2f(y)-f(x) \Rightarrow f(x)f(yf(x))=x^2f(y)$$Then, $P(x,x)$ gives that $f(xf(x))=x^2$. This means that $f$ is surjective for $x>0$. But, as $f(-x)=-f(x)$, so we get that $f$ is in fact always surjective. Now, using injectivity, one can easily prove that $f(1)=1$, and so we have $$P(x,1) \Rightarrow x^2=f(x)f(f(x)) \Longrightarrow f(x)f(yf(x))=x^2f(y)=f(x)f(f(x))f(y) \Rightarrow f(yf(x))=f(f(x))f(y)$$However, as $f$ is surjective, so we can take $f(x)=z$, giving that $f$ is multiplicative also. Let $w=\frac{1}{f(x)}$ (As $f$ is surjective, so $\frac{1}{f}$ is also surjective). Then, using multiplicity, we have $$P \left(y+\frac{1}{f(x)},x \right) \Rightarrow x^2f(y+w)=f(x)f(yf(x)+1)=f(x)(f(yf(x))+1)= x^2f(y)+f(x)$$$$\Rightarrow \text{ As } x^2=f(x)f(f(x)) \text{, we get that } f(f(x))(f(y+w)-f(y))=1$$But, $f(x)=\frac{1}{w}$, and so we get $$f(y+w)-f(y)=\frac{1}{f \left(\frac{1}{w} \right)}=f(w) \Rightarrow f(y+w)=f(y)+f(w)$$where we use that $f \left(\frac{1}{w} \right) f(w)=f(1)=1$. This means that $f$ is both additive and multiplicative, in which case it is well known that $f$ must be the identity function. Hence, done. $\blacksquare$

Nice problem. Here is my solution:

Edit: Just realised that my solution is similar to @2above.

Good problem. Nice solutions. Learned much.

Nice problem Solution $f (x)\equiv 0$ if $f (x)\neq 0$ for some $x $. If $f (x)=f (y) $ then $x=y $ or $x=-y $ $P (0,y)\implies f (0)=0$ or $f (yf (0)-1)=-1\implies f (0)=0$ $P (x,0)\implies f (x)f (-1)=-f (x)\implies f (-1)=-1$ $P (-1,y)\implies -f(-y-1)=f (y)+1$ $\text {Restate the original equation as} $ $f (x)f (-yf (x))=x^2f (y)\implies f (x)f (f (x))=x^2$ $P (x,x): f (-xf (x))=x^2=-f (xf (-x))f (x)+f (-x)=0$ $f (y+1)=f (y)+1 f (x+1)+f (f (x+1))=(x+1)^2$ $f (x)+f (f (x))=2x $, and $f (x)=x $ Hence, $f(x)=0$ $\forall x\in\mathbb R $ and $f(x)=x $. Easy to verify.

Bonus for a TST P3! anantmudgal09 wrote: Find all functions $f: \mathbb{R} \mapsto \mathbb{R}$ such that $$f(x)f\left(yf(x)-1\right)=x^2f(y)-f(x),$$for all $x,y \in \mathbb{R}$. Answers: $f\equiv 0 $ and $f \equiv x \ \forall x \in \mathbb{R}$. Clearly both these functions satisfy, so we continue by proving them to be the only ones. Denote by $(x,y)$ the assertion $f(x)\cdot f(yf(x)-1) = x^2f(y)-f(x)$. $(x,x) \implies f(x)f(xf(x)-1) = f(x)(x^2-1) ... (1)$. Now, $(0,y) \implies f \equiv 0$(which is a solution) or $f(0)=0$. Assuming $f$ to be non-constant, take $f(0)=0$. Now suppose there exists $x_0 \neq 0 \in \mathbb{R}$ such that $f(x_0) =0$. Then, $(x_0,y) \implies x_0^2 \cdot f(y) = 0 \iff x_0 = 0$, contradiction and thus $f$ is injective at $0 \implies f(xf(x) -1) = x^2-1 \implies f$ is surjective over $[-1,\infty)$. Now, $(x,1) \implies f(x) f(f(x)-1) = x^2 - f(x)$. We now show that $f$ is injective. Indeed, if $f(a) =f(b)$, then $a^2 = b^2$. So suppose $a+b=0$. Then we get that $$ f(af(x)-1) = f(-af(x)-1) \forall x \in \mathbb{R}$$. Using surjectivity, pick $\alpha$ such that $f(\alpha) = \frac{1}{a}$. Then we have $f(0)=f(2)$, contradiction $\implies f$ is injective. Now using similar arguments get that $f(x) = -f(-x)$ which implies $f$ is a bijection on $\mathbb{R}$. Then $(x,f(y)) \implies x^2 \cdot f(f(y)) \cdot f(y) = y^2 \cdot f(x) \cdot f(f(x))$ and using $f(1)=1$ (obtainable from $(1,1)$) we establish $f(x) f(f(x)) = x^2$ which in turn implies $f(x+1) = f(x) + 1 \forall x$ and moreover, $f(x) f(yf(x)) = x^2 f(y) = f(x)f(f(x)) f(y) \implies f$ is multiplicative $\implies f(y)f(x+1) = f(y)f(x) + f(y)f(1) \iff f(xy+y) = f(xy) + f(y) \implies f$ is Cauchy and multiplicative and thus $f$ is identity.

$\clubsuit \color{magenta}{\textit{\textbf{ANS:}}}$ $f(x)=0 \quad \textrm{and} \quad f(x)=x \quad \forall x\in \mathbb{R}$. $\blacklozenge \color{blue}{\textit{\textbf{Proof:}}}$ 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). $\quad \blacksquare$

anantmudgal09 wrote: Find all functions $f: \mathbb{R} \mapsto \mathbb{R}$ such that $$f(x)f\left(yf(x)-1\right)=x^2f(y)-f(x),$$for all $x,y \in \mathbb{R}$. Easy, but nice. Let $P(x,y)$ denote the given assertion. $f \equiv 0$ is obviously a solution. Henceforth, assume that $f \not \equiv 0$. Now $$P(1,1): f(1)f(f(1)-1)=0 \implies \exists c \quad \text{such that} f(c)=0$$$$P(c,y): c^2f(y)=0$$since $f \not \equiv 0$, we conclude that $f(t)=0 \iff t=0$. So $f(1)-1=0 \implies f(1)=1$. Observe that $$P(x,1): f(f(x)-1)= \frac{x^2}{f(x)}-1 \quad (i)$$and $$P(1,f(x)): f(f(x)-1)=f(f(x))-1 \quad (ii)$$On comparing $(i)$ and $(ii)$ we have that $f(x)f(f(x))=x^2 \forall x \in \mathbb{R} \quad (iii)$. Now note that for all $x\not = 1,-1$ $$P(x,x): f(xf(x)-1)=x^2-1 \implies f(x^2-1)=f(f(xf(x)-1))=\frac{(xf(x)-1)^2}{f(xf(x)-1)}=\frac{(xf(x)-1)^2}{x^2-1} (iv)$$$$P(f(x),f(x)): f(f(x)f(f(x))-1)= f(x)^2-1 \overset{\text{using} (iii)}{\implies} f(x)^2-1=f(x^2-1)= \frac{(xf(x)-1)^2}{x^2-1}$$Upon simplification, the above result is equivalent to $(f(x)-x)^2=0 \implies f(x)=x$. Also $P(x,0): f(-1)=-1$. Hence we have two solutions, i.e. $$f(x)=x \quad \forall x \in \mathbb{R}$$$$f(x)=0 \quad \forall x \in \mathbb{R}$$Hence, we are done.

$\boxed{f(x)=0}$ works. Assume now that $\exists j:f(j)\ne0$. $P(1,1)\Rightarrow f(1)f(f(1)-1)=0\Rightarrow\exists k:f(k)=0$ $P(k,j)\Rightarrow k^2f(j)=0\Rightarrow k=0$ (injectivity at $0$) Either $f(1)=0$ or $f(f(1)-1)=0$. If $f(1)=0$ then $1=0$, absurd, thus $f(1)-1=0\Rightarrow f(1)=1$. $P(1,x+1)\Rightarrow f(x+1)=f(x)+1$ The assertion becomes $Q(x,y):f(x)f(yf(x))=x^2f(y)$. $Q(x,1)\Rightarrow f(x)f(f(x))=x^2$ $Q(x+1,1)\Rightarrow f(x)+f(f(x))=2x$ So $f(f(x))=2x-f(x)$, thus $f(x)(2x-f(x))=x^2$, which factors as $(x-f(x))^2=0$, hence $\boxed{f(x)=x}$, which is the only remaining solution.

Hello, the equation in hand is f (x) f (y f(x) - 1) = x ^ 2 f (y) - f (x) Assume there is some other function since f (x) =0 x = y = 0 => f (0) f (-1) = -f (0) => f(0) = 0 /f(-1) = -1 Taking f (0) = 0 and substitute y = 0 => f (-1) = -1 Taking f (-1) = -1 and substitute y = 0 => f (0) = 0 This proves that both statements provided before are equivalent Considering that there exists an a which is not equal to 0 such that f (a) = 0, Substituting x = a and y = -1. This will eliminate the case. Substituting P (x, y) P (-1, y) => -f (-y -1) = f(y) + 1 => f (y-1) = -f (-y) -1 Now substitute this in the original equation available to get F (x) f (f (x)) = x^2 P(1, 1) f(1) f (f(1) -1) = 0 f(1) = 1 Then, P (1,y) = f; This is an odd function Use f (x +1) = f(x) +1 to get rid of the 1 Put x = x + 1, y = 1 to get F (f(x)) = 2 x - f (x) Now substitute f (x) f (f(x)) = x ^2 with f (f (x)) = 2x - f (x) (f(x) - x) ^ 2 = 0 f(x) = x Working functions: f (x) = 0 and f (x) =x

We will prove that $f\equiv0$ and $f(x)=x$ are the only solutions. Let $P(x,y)$ be the assertion in $f(x)f\left(yf(x)-1\right)=x^2f(y)-f(x)$. It can be easily seen that the only constant function that works is $\boxed{f \equiv 0}$, which is our first solution. Let $f$ be non-constant. Claim 1 : $f(0)=0$. Proof : $P(0,y) : f(0)f(yf(0)-1)=-f(0)$ Assume possible $f(0)\neq0$. Then, $f(yf(0)-1)=-1$. As $f(0)\neq0$, as we vary $y$ over $\mathbb{R}$, $yf(0)-1$ also varies over $\mathbb{R}$. Hence, $f(x)=-1 \forall x $.But $f\equiv-1$ doesn't satisfy given equation. Contradiction! Hence proved claim 1! Claim 2 : $f$ injective. Proof : Let $f(a)=f(b)$. Let $c$ be such that $f(c)\neq0$. $P(a,c)-P(b,c) : a^2=b^2 \implies a = \pm b$. Hence, $f(a)=f(b) \implies a= \pm b ...(1)$. Assume that $a=-b$ ($a,b \neq 0$) is possible. $P(c,a)-P(c,b) : f(af(c)-1)=f(bf(c)-1) \overset{(1)}{\implies} af(c)-1= \pm (bf(c)-1) \overset{a=-b}{\implies} af(c)-1= \pm (-af(c)-1)$. If $af(c)-1=-af(c)-1$, we get $a=0$. Contradiction. If $af(c)-1=-(-af(c)-1)$, we get $-1=1$. Contradiction! Hence, if $f(a)=f(b)$, $a=-b$ is not possible. Combining with $(1)$, we get $f(a)=f(b)\implies a=b$. Hence proved claim 2! Hence note that , $f(x) \neq 0 \forall x\neq 0$. Claim 3 : $f(1)=1$. Proof : $P(1,1) : f(1)\cdot f(f(1)-1)=0 \implies f(f(1)-1)=0 \implies f(1)-1=0 \implies f(1)=1$. (Here we used claim 1 and claim 2) Hence Proved claim 3! Claim 4 : $f(-1)=-1$. Proof : $P(x,0) : f(x)f(-1)=-f(x) \implies f(-1)=-1$. (Here we used claim 1 and claim 2) Hence proved claim 4! Claim 5 : The given equation reduces to $f(x)f(yf(x))=x^2f(y)$. Proof : $P(1,y) : f(y-1)=f(y)-1...(2)$. (Here we used claim 3) $y\rightarrow yf(x) : f(yf(x)-1)=f(yf(x))-1$. Substituting this in given equation, we get $$f(x)f(yf(x))=x^2f(y)$$. Hence proved claim 4! Let $R(x,y)$ be the assertion in $f(x)f(yf(x))=x^2f(y)$. Claim 6 : $f(x+1)=f(x)+1$ and $f(x^2)=f(x)^2$. Proof : $(2) : f(y)=f(y-1)+1 \implies f(y+1)=f(y)+1$. $R(f(x),f(x)) : f(f(x)f(f(x)))=f(x)^2$. $R(x,1) : f(x)f(f(x))=x^2 \implies f(f(x)f(f(x))=f(x^2)$. Hence, $f(x)^2=f(x^2)$ (From last two equations). Hence proved claim 6! From claim 6, its well-known that ${f(x)=x}$ is the solution. $\boxed{f(x)=x}$ indeed works and is our second solution! Hence, we are done

Let $P(x,y)$ denote the given assertion. $P(0,x): f(0)f(xf(0)-1)=-f(0)$. So either $f(0)=0$ or $f(xf(0)-1)=-1$ for any $x$. If $f(0)\ne 0$, then $xf(0)-1$ can take on any real value, which implies $f\equiv -1$, which is not a solution. Thus, we have $f(0)=0$. Now, noting $\boxed{f\equiv 0}$ works, we can assume $f$ is non-constant. Claim: $f$ is injective. Proof: Suppose $f(a)=f(b)$ with $a\ne b$. $P(a,x): f(a)f(xf(a)-1)=a^2f(x)-f(a)$. $P(b,x): f(b)f(xf(b)-1)=b^2f(x)-f(b)$. This implies $a^2f(x)=b^2f(x)$. If we set $x$ such that $f(x)\ne 0$, then $a^2=b^2\implies a=\pm b\implies a=-b$, since $a\ne b$. Then $f(a)=f(-a)$. In fact, this implies $f$ is injective at $0$. $P(a,-a): f(a)f(-af(a)-1)=a^2f(a)-f(a)$. $P(-a,a): f(a)f(af(a)-1)=a^2f(a)-f(a)$. If $f(a)\ne 0$, then we have $f(-af(a)-1)=f(af(a)-1)$. However, this implies either $af(a)+1=af(a)-1$, or $-af(a)-1=af(a)-1$, both are absurd. So $f(a)\ne f(-a)$. If $f(a)=0$, then $a=0$. Since $f$ is injective at $0$, $f$ is injective. $\blacksquare$ $P(x,0): f(x)f(-1)=-f(x)$. If we set $x\ne 0$, then we get $f(-1)=-1$. $P(1,1): f(1)f(f(1)-1)=0$, so $f(1)=1$. $P(1,x): f(x-1)=f(x)-1\implies f(x+1)=f(x)+1$. Now we rearrange the FE. We have $f(x)(f(yf(x))-1)=x^2f(y)-f(x)$, so \[f(x)f(yf(x))=x^2f(y)\] Let $Q(x,y)$ be the assertion here. $Q(x,1): f(x)f(f(x))=x^2\implies f(f(x))=\frac{x^2}{f(x)}$. $P(x+1,1): (f(x)+1)f(f(x))=(x+1)^2-f(x)-1=x^2+2x-f(x)\implies f(f(x))=\frac{x^2+2x-f(x)}{f(x)+1}$. This implies\begin{align*} \frac{x^2}{f(x)}=\frac{x^2+2x-f(x)}{f(x)+1} \\ \implies x^2(f(x)+1)=f(x)(x^2+2x-f(x)) \\ \implies x^2f(x)+x^2=x^2f(x)+2xf(x)-f(x)^2 \\ \implies x^2=2xf(x)-f(x)^2 \\ \implies f(x)^2-2xf(x)+x^2=0 \\ \implies (f(x)-x)^2=0 \\ \implies \boxed{f(x)=x} \\ \end{align*}which clearly works.

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The answer is $f(x)=x$ and $f \equiv 0$ only, which both work. Hence suppose $f \not \equiv 0$. Let $P(x,y)$ denote the assertion. From $P(x,0)$ we have $f(x)(f(-1)+1)=x^2f(0)$. If $f(-1)+1 \neq 0$, then $f(x)=cx^2$ for some nonzero constant $c \in \mathbb{R}$. But this clearly doesn't work, so $f(-1)=-1$ and thus $f(0)=0$. Then from $P(-1,y)$ we obtain $f(-y-1)=-f(y)-1$. If $f(x)=0$ for some $x$, by picking $y$ such that $f(y) \neq 0$ we find that $x=0$. Then for $x \neq 0$, $P(x,x)$ gives $f(xf(x)-1)=x^2-1$, hence the range of $f$ contains $(-1,\infty)$. Since $f(-y-1)=-f(y)-1$, the range of $f$ also contains $(-\infty,1)-1$, so $f$ is surjective. Additionally, $P(-1,-1)$ implies $f(-2)=-2$, hence $-f(1)-1=f(-2) \implies f(1)=1$. Using $f(-y-1)=-f(y)-1$, from $P(x,-1)$ we obtain $f(x)(f(-f(x)-1)+1)=-x^2 \implies f(x)f(f(x))=x^2$. Thus we can rewrite the assertion as $f(yf(x)-1)=f(f(x))f(y)$. From surjectivity, this becomes $f(xy-1)=f(x)f(y)-1$. Comparing $(x,y)$ with $(xy,1)$ and using $f(1)=1$ implies that $f$ is multiplicative. On the other hand, by plugging in $y=1$ we also get $f(x-1)=f(x)-1$. Since $f$ is multiplicative, we have $f(x^2)=f(x)^2>0$ for all $x>0$, hence $f$ sends positive reals to positive reals. Thus let $g(x)=\log f(e^x)$ sending real numbers to real numbers, so $g$ is additive (by only considering the multiplicativity of $f$ over $\mathbb{R}^+$). On the other hand, $f(x-1)=f(x)-1$ implies that $f(x)>1$ for $x>1$, hence $g(x)>0$ for $x>0$. Therefore $g$ is linear, so $f(x)=x^k$ for some $k \in \mathbb{R}$. Then we have $2^k-1=1$, hence $k=1$ and $f(x)=x$ as desired. $\blacksquare$