Let $P(x,y)$ be assertion. $P(0,0)$ gives $f(0)=f(0)^2$, $f(0)=0$ or $1$ if $f(0)=1, P(0,y) $ gives $f(y)=1$ if $f(0)=0$, $P(x,0)$ gives $f(x^2)=f(x)^2$ and $P(-x,x)$ gives $f(x^2)=xf(x)$ $\therefore xf(x)=f(x)^2$, $f(x)=x$ or $f(x)=0$ therefore the 3 solutions are $f(x)=0,f(x)=1,f(x)=x$

qwerty_ytrewq wrote: $\therefore xf(x)=f(x)^2$, $f(x)=x$ or $f(x)=0$ therefore the 3 solutions are $f(x)=0,f(x)=1,f(x)=x$ You should also check that there can't be some values of $a$ and $b$ such that $f(a) = 0$, but $f(b) = b$. i.e. $f(x)^2 = xf(x)$ tells us that for any particular value of $x$, we have that $f(x) = x$ or $f(x) = 0$, but it doesn't guarantee that we have the same option for every $x$.

qwerty_ytrewq wrote: Let $P(x,y)$ be assertion. $P(0,0)$ gives $f(0)=f(0)^2$, $f(0)=0$ or $1$ if $f(0)=1, P(0,y) $ gives $f(y)=1$ if $f(0)=0$, $P(x,0)$ gives $f(x^2)=f(x)^2$ and $P(-x,x)$ gives $f(x^2)=xf(x)$ $\therefore xf(x)=f(x)^2$, $f(x)=x$ or $f(x)=0$ therefore the 3 solutions are $f(x)=0,f(x)=1,f(x)=x$ pointwise trap

DylanN wrote: qwerty_ytrewq wrote: $\therefore xf(x)=f(x)^2$, $f(x)=x$ or $f(x)=0$ therefore the 3 solutions are $f(x)=0,f(x)=1,f(x)=x$ You should also check that there can't be some values of $a$ and $b$ such that $f(a) = 0$, but $f(b) = b$. i.e. $f(x)^2 = xf(x)$ tells us that for any particular value of $x$, we have that $f(x) = x$ or $f(x) = 0$, but it doesn't guarantee that we have the same option for every $x$. i see, bit of a dummy here, let me complete the solution, suppose that there exist non-zero value $x,y$ such that $f(x)=0$ and $f(y)=y$ $f(x^2)=f(x)^2=0$ then $P(x,y)$ gives $f(x+y)=y$ which does not equal to $0$ or $x+y$, and gives a contradiction. $\therefore f(x)=0,f(x)=1,f(x)=x$. hope there is no mistakes now

$P(x,y)$ be the assertion of $f(x^2)-yf(y)=f(x+y)(f(x)-y)$ $P(0,0)$ $f(0)=0$ or $f(0)=1$ $P(0,x)$:[when $f(0)=1$] $f(x)=1$, which is a solution. $P(-x,x)$[when $f(0)=0)$] $f(x^2)=xf(x)$ $P(x,0)$: $f(x^2)=f(x)^2$ So, $f(x)^2=xf(x)$ $f(x)=0 $ or $f(x)=x$ Now, let $f(a)=0 $ and $f(b)=b$ where a,b are nonzero real numbers. $P(a,b)$: $f(a)^2-bf(b)=f(a+b)(f(a)-b)$ $-b^2=f(a+b)(-b)$ $b=f(a+b)$ When a=0. But we told a,b are nonzero. So, a contradiction. Hence, $f(x)=0,1,x$ Is this right?

$$P(x,y)\Rightarrow f(x^2)-yf(y)=f(x+y)(f(x)-y)$$$$P(0,0)\Rightarrow f(0)=f(0)^2\Rightarrow f(0)=\{1,0\}$$$$P(0,x)\Rightarrow f(x)=1$$And that is a solution, now $f(0)=0$ Take $c$ such that $f(c)=0$ then $$P(x,0)\Rightarrow f(x^2)=f(x)(f(x))=f(x)^2$$$$P(x,-x)\Rightarrow f(x^2)+xf(-x)=0$$And for $x\neq0$ we have $f(-x)=-f(x)$ Then $f(x)(f(x)-x)=0$ for all $x\neq0$ then if $f(x)\neq 0\Rightarrow f(x)=x$ Suppose that exist $x$ such that $f(x)=0$ and $x\neq0$ then: $$P(c,y)\Rightarrow yf(y)=f(y+c)y\Rightarrow f(y)=f(y+c)$$take $y$ such that $f(y)=y$ and $y\neq 0$ then $f(y+c)=y$ then $y=0$ or $y=y+c$ then $y=y+c$ and that implies $c=0$. That implies $f(x)=0\iff x=0$ therefore $f(x)=x\ x \in\ \mathbb{R}$ is another solution. Solutions: $f(x)=x$, $f(x)=1$, $f(x)=0$

Ucchash wrote: $P(x,y)$ be the assertion of $f(x^2)-yf(y)=f(x+y)(f(x)-y)$ $P(0,0)$ $f(0)=0$ or $f(0)=1$ $P(0,x)$:[when $f(0)=1$] $f(x)=1$, which is a solution. $P(-x,x)$[when $f(0)=0)$] $f(x^2)=xf(x)$ $P(x,0)$: $f(x^2)=f(x)^2$ So, $f(x)^2=xf(x)$ $f(x)=0 $ or $f(x)=x$ Now, let $f(a)=0 $ and $f(b)=b$ where a,b are nonzero real numbers. $P(a,b)$: $f(a)^2-bf(b)=f(a+b)(f(a)-b)$ $-b^2=f(a+b)(-b)$ $b=f(a+b)$ When a=0. But we told a,b are nonzero. So, a contradiction. Hence, $f(x)=0,1,x$ Is this right? $ f $ can't be periodic?

Clearly if $f$ is constant then its $0,1$ so we will assume that $f$ is non-constant. Let $P(x,y)$ the assertion of the given F.E. $P(x,0)$ $$f(x^2)=f(x)^2 \overset{x=0}{\implies} f(0)^2=f(0) \implies f(0)=0 \; \text{or} \; f(0)=1$$If $f(0)=1$ then do $P(0,x)$ $$1-xf(x)=f(x)-xf(x) \implies f(x)=1 \; \text{contradiction!!}$$Hence $f(0)=0$. Now assume that there exists $c \ne 0$ such that $f(c)=0$, then by $P(c,x)$ $$f(x)=f(c+x) \implies f \; \text{periodic with period} \; c$$Now do $P(x,c)$ $$f(x)=0 \; \text{contradiction!!}$$Hence $f$ is injective at $0$ so by the first formula we have $f(1)=1$ $P(x,f(x))$ $$f(f(x))=f(x)$$Now we prove general injectivity. Let $f(a)=f(b)$ and assume $a \ne b$ then $P(a,x-a)-P(b,x-a)$ for $x \ne f(a)+a=f(b)+a$ $$f(x)=f(x+b-a)$$Now let $x \ne f(a)+a=f(b)+a$ and do $P(x,b-a)$ $$f(x)=f(b-a)$$So $f$ is constant everywhere except at 1 fixed point. Set $P(a+f(a),x)$ where $x \ne 0$, clearly $f(b-a) \ne 0$. $$f(a+f(a))=1 \implies f(b-a)=1 \implies f(x)=1 \; \text{contradiction!!}$$Hence $f$ must be injective and using that on $f(f(x))=f(x)$ we get that $f(x)=x$. Hence the solutions are $\boxed{f(x)=0,1,x \; \forall x \in \mathbb R}$ thus we are done

Solved with Taco12. We claim that the functions $f(x) = 0, f(x) = 1,$ and $f(x) = x$ are the only solutions that satisfy the equation. Let $P(x, y)$ denote the given assertion. $P(x, 0)$: $f(x^2) = f(x)^2$. $P(0, y)$: $f(0) - yf(y) = f(y) \cdot (f(0) - y) \Rightarrow f(0) - yf(y) = f(y)f(0) - yf(y) \Rightarrow f(y)f(0) - f(0) = 0 \Rightarrow f(0)(f(y) - 1)$. This means that either $f(0) = 0$ or $f(x) = 1$. Plugging the function $\boxed{f(x) = 1}$ into the original equation, we see that it works. Henceforth, we shall assume $f(0) = 0$. $P(-y, y)$: $f(y^2) - yf(y) = f(0)f(-y)-y = 0 \Rightarrow f(y)^2 - yf(y) = 0 \Rightarrow f(y)(f(y)-y) = 0$. This results in two more possible solutions: $\boxed{f(x) = 0}$ and $\boxed{f(x) = x}$. Plugging these in, we see that they work. We are done. $\blacksquare$