danepale wrote: Let $\mathbb{R}$ denote the set of the reals. Find all $f : \mathbb{R} \to \mathbb{R}$ such that $$ f(x)f(y) = xf(f(y-x)) + xf(2x) + f(x^2) $$for all real $x, y$. $\boxed{\text{S1 : }f(x)=0\text{ }\forall x}$ is a solution. So let us from now look only for non allzero solutions. Let $P(x,y)$ be the assertion $f(x)f(y)=xf(f(y-x))+xf(2x)+f(x^2)$ $P(0,x)$ $\implies$ $f(0)(f(x)-1)=0$ and since $f\equiv 1$ is not a solution, we get $f(0)=0$ Subtracting $P(x,x)$ from $P(x,y)$, we get new assertion $Q(x,y)$ : $f(x)(f(y)-f(x))=xf(f(y-x))$ If $f(u)=0$ for some $u\ne 0$, then $Q(u,x+u)$ $\implies$ $f(f(x))=0$ $\forall x$ Then $Q(x,0)$ $\implies$ $f(x)=0$ $\forall x$, impossible. So $f(x)=0$ $\iff$ $x=0$ and $Q(x,y)$ implies that $f(x)$ is injective. $Q(x-y,0)$ $\implies$ $f(f(y-x))=-\frac{f(x-y)^2}{x-y}$ $\forall x\ne y$ And $Q(x,y)$ becomes $f(x)(f(y)-f(x))=-\frac x{x-y}f(x-y)^2$ $\forall x\ne y$ Switching $x,y$ and dividing, we get $\frac{yf(x)}{xf(y)}=\frac{f(x-y)^2}{f(y-x)^2}$ $\forall x,yx-y\ne 0$ This implies that $g(x)=\frac{f(x)}x$, from $\mathbb R\setminus\{0\}\to\mathbb R\setminus\{0\}$ matches : $\frac{g(x)}{g(y)}=\frac{g(x-y)^2}{g(y-x)^2}$ We conclude that $g(x)$ is nonzero and of constant sign and equation $\frac{g(x)}{g(y)}=h(x-y)$ gives then easily result $g(x)=\alpha e^{a(x)}$ for some additive $a(x)$ Plugging this in $\frac{g(x)}{g(y)}=\frac{g(x-y)^2}{g(y-x)^2}$, we get $a(x)=0$ and $g(x)$ constant. Plugging this back in original equation, this becomes $\boxed{\text{S2 : }f(x)=3x\text{ }\forall x}$

Well,maybe this is too late,but having more solution is better,right?

Here's my solution (hopefully correct): Let $x=y=0$, then $f(0)^2=f(0)$, so that means $f(0)=0$ or $f(0)=1$. If $f(0)=1$, then $x=0$ gives us $f(0)f(y)=f(0)$ or $f(y)=1$, but it doesn't satisfy the initial equation. So $f(0)=0$. We set $y=0$ and $y=x$ to get $$0=xf(f(-x))+xf(2x)+f(x^2),$$$$f(x)^2=xf(2x)+f(x^2),$$so combining these we get $f(x)^2=-xf(f(-x))$. Now, $x=y=1$ gives us $f(1)^2=f(1)+f(2)$ and $x=1$ gives us $$f(1)f(y)=f(f(y-1))+f(2)+f(1),$$so denoting $f(1)=c$, we get $$cf(x)=f(f(x-1))+c^2,$$or $$f(f(x))=cf(x+1)-c^2.$$Now using that in the initial equation, we get $$f(x)f(y)=x(cf(y-x+1)-c^2)+xf(2x)+f(x^2),$$$$f(x)f(y)=cxf(y-x+1)-xc^2+xf(2x)+f(x^2),$$and switching $y$ by $y+x$ gives us $$f(x)f(y+x)=cxf(y+1)-xc^2+xf(2x)+f(x^2),$$plugging $x=2$ gives $$f(2)f(y+2)=2cf(y+1)-2c^2+3f(4),$$(if $c=0$, then $f$ is a constant, checking gives $f(x)=0$ which satisfies, $c=1$ gives also a constant function) $$(c^2-c)f(x+1)=constantf(x)+3f(4)-2c^2,$$$$f(x+1)=constantf(x)+\frac{3f(4)-2c^2}{c^2-c}=constantf(x)+k,$$but we know that $$f(f(x))=cf(x+1)-c^2=constantf(x)+b=af(x)+b$$for real constants $a$ and $b$. Now using that we plug that into the initial equation: $$f(x)f(y)=x(af(y-x)+b)+f(x)^2,$$using $y=x$ gives us $xb=0$ for all real numbers $x$, which is only true if $b=0$, so that means $f(f(x))=af(x)$. But we know that $f(x)^2=-xf(f(-x))$ or in other words $f(x)^2=-xaf(-x)$, but switching $x$ by $-x$ gives us $f(-x)^2=xaf(x)$, squaring the last one and some algebraic manipulations gives $$f(x)^4=x^2a^2f(-x)^2=x^2a^2xaf(x)=x^3a^3f(x),$$so we get that $f(x)=0$ for some $x$ or $f(x)= ax$ for some $x$, where $a$ is a nonzero constant. Now, $f(1)=0$ or $f(1)=a$. If the first thing is the case, then $f(f(x))=cf(x+1)-c^2=0$, because remember $f(1)=c$, so $af(x)=0$ for all real $x$, that means $f(x)=0$, which satisfies the initial equation. Now $f(1)=a$. We can obtain the following equation by switching $y$ by $y+x$: $$f(x)f(y+x)=xf(f(y))+f(x)^2.$$But if for some nonzero $k$ we have $f(k)=0$, then plugging $x=k$ gives us $kf(f(y))=0$, so $af(y)=0$, so $f(y)=0$ for all real $y$, which we don't want, so $k=0$. But $f(0)=0$ satisfies the equation $f(x)=ax$, so $f(x)=ax$ is also a solution, but verification gives us $a=3$, so $f(x)=3x$.

Can someone give an elementary solution to this problem?