My progress: If we put $x=y=0$, we get $f(f(0))=0$ If we put just $x=0$, we get $f(f(y))=f(y)+(y-1)f(0)$ Now put in this $y\to x+y$, and we get $f(f(x+y))=f(x+y)+(x+y-1)f(0)$. And on the basis of the proposed problem we get: $f(2xy)+(x+y-1)f(0)=xf(y)+yf(x)$. Putting $y=1/2$, we get that $f(x)=ax+b$, and this must be introduced to the oroginal equation.

TuZo wrote: $f(2xy)+(x+y-1)f(0)=xf(y)+yf(x)$. Putting $x=y=0$, we get $f(0)=0$. This part is wrong, you get $0=0$. But you can finish from there just take $y=\frac{1}{2}$ and you find that $f(x)=ax+b$ .

Take $x=y=0$ to get $f(f(0))=0$, and now assert $(f(0), 0)$ to get $2f(0)=f(0)^2$ so we get two cases: $1) \text{ } f(0)=2$ EDIT: My original solution had a mistake, but when resolving the problem the solution ended up being exactly the same as the second case and every other solution so it would be dumb if I posted it $2) \text{ } f(0) = 0$ Plugging in $(x, 0)$ gives $f(f(x))=x$, so the equation reduces to $f(2xy) = xf(y)+yf(x)$, but taking $y=\frac{1}{2}$ gives $f(x)= 2f(\frac{1}{2})x$ so $f$ is linear $\blacksquare$

$P(x,0): f(f(x))=f(x)+xf(0)-f(0)$. Thus $P(x,y): f(2xy)+(x+y-1)f(0)=xf(y)+yf(x)$, using $f(f(x+y))=f(x+y)+(x+y-1)f(0)$. $P(\frac 12,y): \frac{f(y)}2=y(f(\frac 12)-f(0))+\frac 12 f(0) \implies f(y)=ay+b$. Putting $f(x)=ax+b$ in original expression, we get $(a^2-(a+b)x)+(a^2-(a+b)y)+b(a+1)=0$. Case 1: $a=-1 \implies b=2$. Case 2: $b=0 \implies a=1$ or $0$. Thus our 3 solutions are $f(x)=2-x$, $f(x)=x$ and $ f \equiv 0$.

Wow it's easier than I expected it to be. $P(x,0)\implies f(f(x))=f(0)x+f(x)-f(0)$ So, the original equation changes to $f(2xy)+(x+y)f(0)=xf(y)+yf(x)+f(0)$ Let $f(0)=b$ Let $g(x)=f(x)-b$ Then, the equation transforms to $g(2xy)=yg(x)+xg(y)$ Let $g\left(\frac12\right)=a$ $P\left(x,\frac12\right)\implies g(x)=2ax$ So, $f(x)=2ax+b$. Substituting this in the original equation, we find that $f(x)=2-x, f(x)=x$ and $f(x)\equiv 0$ are the only solutions . Indeed, Q.E.D.

dangerousliri wrote: TuZo wrote: $f(2xy)+(x+y-1)f(0)=xf(y)+yf(x)$. Putting $x=y=0$, we get $f(0)=0$. This part is wrong, you get $0=0$. But you can finish from there just take $y=\frac{1}{2}$ and you find that $f(x)=ax+b$ . Yes, thank you!

niksic wrote: Take $x=y=0$ to get $f(f(0))=0$, and now assert $(f(0), 0)$ to get $2f(0)=f(0)^2$ so we get two cases: $1) \text{ } f(0)=2$ EDIT: My original solution had a mistake, but when resolving the problem the solution ended up being exactly the same as the second case and every other solution so it would be dumb if I posted it $2) \text{ } f(0) = 0$ Plugging in $(x, 0)$ gives $f(f(x))=x$, so the equation reduces to $f(2xy) = xf(y)+yf(x)$, but taking $y=\frac{1}{2}$ gives $f(x)= 2f(\frac{1}{2})x$ so $f$ is linear $\blacksquare$ Wow man you just forgot the case $f(0) = 2$.

Posting for fun. Let $P(x,y)$ be the assertion. $P(0,0)\implies f(f(0))=0$. $P(x,0)\implies f(0)+f(f(x))=f(x)+xf(0)$. If $f(0)=0$, then the given FE takes the form, $f(2xy)=xf(y)+yf(x)$. Taking $y=\frac{1}{2}$ to the last equation, we get that $f(x)=2xf\left(\frac{1}{2}\right)$. Plugging $f(x)=cx$ back into our original FE, we get that $c=0$ or $c=1$. Now, if $f(0)\neq 0$, then $f$ is injective, hence, $P(1,0)\implies f(f(1))=f(1)\implies f(1)=1$. $P(1,1)\implies f(2)+f(f(2))=2f(1)+f(2)\implies f(f(2))=2$. $P(2,0)\implies f(2)+f(0)=2$ $P(f(2),0)\implies 2=f(0)+f(2)=f(2)f(0)+2\implies f(2)f(0)=0\implies f(2)=0$ as we assumed that $f(0)\neq 0$. Therefore, $f(0)=2$. Hence, we have $2+f(f(x))=f(x)+2x$. Furthermore, our original equation takes the following form: $f(2xy)+2(x+y)-2=xf(y)+yf(x)$. Setting $y=\frac{1}{2}$, then we get that $f(x)=kx+2$ for some constant $k$. Plugging this back into our original, then $2+2k=0$ and $k^2(x+y)=(k+2)(x+y)$, thus $k=-1$. Answer. $f(x)=0\forall x\in\mathbb R$; $f(x)=x\forall x\in\mathbb R$ and $f(x)=2-x\forall x\in\mathbb R$.

Case 1: $f(0)\ne0$ We have $f(f(x))=f(x)+2x-2$ by $P(x,0)$, since $f(0)$ must be $2$. So $P(x,y)$ becomes $f(2xy)+2x+2y-2=xf(y)+yf(x)$. Let $g(x)=f(x)-2$. $Q(x,y):g(2xy)=xg(y)+yg(x)$ $Q\left(x,\frac12\right)\Rightarrow g(x)=2xg\left(\frac12\right)$ Testing, we see that the solution is $\boxed{f(x)=-x+2}$. Case 2: $f(0)=0$ $P(x,0)\Rightarrow f(f(x))=f(x)$ It reduces to $f(2xy)=xf(y)+yf(x)$, same as above, except with the additional linear solutions $\boxed{f(x)=0}$ and $\boxed{f(x)=x}$.

Let $P(x,y)$ be the assertion. $P(0,0)\implies f(f(0))=0$. $P(x,0)\implies f(f(x)) = f(x) + xf(0) - f(0)$. Plugging in $x=f(0)$ in above, we get $2f(0)=f(0)^2$ Case 1: $f(0)=0$ Then we get $f(f(x))=f(x)$ for all $x\in \mathbb{R}$. Original equation rewrites as $f(2xy) = xf(y) + yf(x)$. Plugging in $y=\frac{1}{2}$ in above, we get $f(x)=cx$ for some constant $c$ and plugging in the original equation we see that $c=0$ or $c=1$ so we get $\boxed{ f(x)=x \forall x\in \mathbb{R}}$ and $\boxed{ f(x)=0 \forall x\in \mathbb{R}}$ as solutions. Case 2: $f(0)=2$ Then we get $f(f(x)) = f(x) + 2x - 2$ for all $x\in \mathbb{R}$. Original equation rewrites as $f(2xy) + 2(x+y) - 2 = xf(y) + yf(x)$. Plugging in $y=\frac{1}{2}$ in above, we get $f(x)=cx + 2$ for some constant $c$ and plugging in the original equation we see that $c=-1$ so we get $\boxed{ f(x)=2-x \forall x\in \mathbb{R}}$.

Let $P(x, y)$ denote the assertion. \begin{align*} P(0, 0) &\implies f(f(0)) = 0 \\ P(f(0), 0) &\implies 2f(0)=f(0)^ 2 \implies f(0) = 0, f(0)=2 \end{align*}$\textbf{Case 1:}$ $f(0) = 0$ We have $f(f(x)) = f(x)$ which means that \[ f(2xy) = xf(y) + yf(x) \]Substituting $y=\frac{1}{2}$ we get $f(x)=cx$. Then substituting this we get that $c=0$ or $c=1$. This means that $f(x)=0$ or $f(x)=x$ are functions that satisfy the equation. $\textbf{Case 2:}$ $f(0) = 2$ We have $f(f(x)) = f(x) + 2x - 2$ which means that \[ f(2xy) + 2x + 2y = xf(y) + yf(x) + 2. \]Again plugging $y=\frac12$ we get \[ f(x) + 2x + 1 = xf(1/2) + 1/2f(x) + 2 \implies f(x) = cx + 2 \]Solving we find that $c=-1$. Thus we also have the function $f(x) = 2 - x$.

All linear functions are $x\mapsto 0,$ $x\mapsto x$ and $x\mapsto 2-x.$ Denote the assertion by $P(x,y).$ We prove linearity. $P(x+y,0)\implies f(f(x+y))=f(x+y)+(x+y-1)f(0)$ $P(x,y)\implies f(2xy)+(x+y-1)f(0)=xf(y)+yf(x)$ $P(x,1/2)\implies f(x)=mx+n$