Let $\rm P(x,y): f(xy+f(xy))=2xf(y)$. $\rm P(0,0): f(f(0))=0 \,\,(1),\,\,\,P(x,0): f(f(0))=2xf(0)\overset {(1)}{\Leftrightarrow }0=2xf(0)\Leftrightarrow \boxed {\rm f(0)=0}$. $\rm P(x,1): f(x+f(x))=2xf(1)\,\,(2)\,\,, P(1,x): f(x+f(x))=2f(x)\,\,(3)$ $\rm (3)\overset{(2)}{\Rightarrow} 2f(x)=2xf(1)\Leftrightarrow \boxed {\rm f(x)=xf(1)}$ So the original $\rm f(1)(xy+f(1)xy)=2xf(1)y$.If $\rm f(1)=0$ then $\boxed {\rm f(x)=0 \,\,\,,\forall x\in \mathbb{R}}$ . If $\rm f(1) \neq 0$ then $\rm xy+f(1)xy=2xy\Leftrightarrow f(1)xy=xy,\forall x,y \in \mathbb{R} $ so take $\rm x,y \neq 0$ ,we have that $\rm f(1)=1$. So the solutions are $\rm f(x)=x,f(x)=0,x\in \mathbb{R}$

If we let $f(1)=c$ from $P(1,x)-P(x,1)$ we get $f(x)=cx$. And checking we get that the solutions are $f(x)=0$ and $f(x)=x$

parmenides51 wrote: Find all functions $f : R \to R$ that satisfy $f (xy + f(xy)) = 2x f(y)$ for all $x, y \in R$ Interesting. Let $P(x, y)$ be the assertion. We claim that the two solutions to the given functional equation are $\boxed{f(x)=x}$ for all reals $x$ and $\boxed{f(x)=0}$ for all reals $x$. It can be easily verified that these two solutions work, now it suffices to show that these are the only solutions to the given functional equation. $P(0, 0) \implies f(f(0))=0$ and $P(x, 0) \implies f(f(0))=2xf(0)$ and these two assertions put together implies that $\boxed{f(0)=0}$. Now let $k \neq 0$. Then $P\left ( \dfrac{1}{k}, k\right ) \implies f(k)=ck$ where $c$ is a constant and I leave it upto the reader to see why this assertion is true. Substituting $f(k)=ck$ back into the equation gives us two solutions namey $\boxed{f(x)=x}$ for all reals $x$ and $\boxed{f(x)=0}$ for all reals $x$ as desired.

We claim that $f(x)=x$ and $f(x)=0$ are the only two solutions. It's easy to check that they works. Now we show these are the only ones By $P(xy,1)$ conclude $2xyf(1)=2xf(y)$. Hence $f(x)=xf(1)~\forall x\in\mathbb{R}$. Substitute back in $P$ to get $$f(x)=0~\forall x\in\mathbb{R}~~~\text{and}~~~f(x)=x~\forall x\in\mathbb{R}$$as desired.

For the solution bank. Let $P(x,y)$ be the insertion of given functional equation. $P(x,y)\implies f (xy + f(xy)) = 2x f(y)$ $P(y,x)\implies f (xy + f(xy)) = 2y f(x)$ Hence, $x f(y)=y f(x)$ If $x$ or $y$ is $0$, then we get that $f(0)=0$, so $\frac{f(x)}{x}=\frac{f(y)}{y}\forall x,y\neq 0\in \mathbb R$, hence $f(x)=cx$. Plugging this back into our given equation, we obtain that $c^2xy+cxy=2cxy\implies c=0\text{ or } 1$. These indeed work. Answer. $\boxed{f(x)=x, f(x)=0}.$

Let $P(x,y)$ be the assertion $f (xy + f(xy)) = 2x f(y)$. Comparing $P(y,x)$ with $P(x,y)$, we get $Q(x,y):2xf(y)=2yf(x)$, so $\frac{f(x)}x=\frac{f(y)}y$ and $\frac{f(x)}x$ is constant. Thus, $f(x)=cx\forall x\ne0$ for a constant $c$. Now $Q\left(0,\frac12\right)\implies f(0)=0$, and we see that the only solutions that work are $\boxed{f\equiv0}$ and $\boxed{f\equiv x}$.

Solution with arpan_the_gad** and jelena_ivanchic $P(x,0): f(f(0))=xf(0)\implies f(0)=0$. Now for non-zero $x$ and $y$ we have $P(x,y)$ and $P(y,x)\implies yf(x)=xf(y)\implies f(x)/x=f(y)/y=c\implies f(x)=cx$ for all $x\neq 0$. Now left as an exercise to the reader

Let $P(x, y)$ be the given assertion. \begin{align*} P(x,x) &\implies f(x^2 + f(x^2)) = 2xf(x), \\ P(0, x) &\implies f(f(0)) = 0,\\ P(1, x) &\implies f(x + f(x)) = 2f(x) [1],\\ P(x, 0) &\implies f(f(0)) = 2xf(0) = 0,\\ P(x, 1) &\implies f(x+f(x)) = 2xf(1) [2]. \end{align*}Note that from $[1]$ and $[2]$ we get: $$2f(x) = 2xf(1),$$$$f(x) = xf(1).$$Let $f(1) = c \in \mathbb{R}.$ Assume $c\neq 0$. Plugging back into our original equation: \begin{align*} f(xy + f(xy)) &= 2xf(y),\\ f(xy + cxy) &= 2xyc,\\ cxy + c^2xy &= 2xyc,\\ c^2xy &= cxy,\\ cxy &= xy. \end{align*}Thus the only possible value of $c$ is $1$. This gives us our first solution $f(x) = x\forall x\in \mathbb{R}$. Let $f(1) = c = 0$, then $f(x) = 0 \forall x\in \mathbb{R}$. This solution clearly works. Thus the only two solutions are: $$f(x) = x$$and $$f(x) = 0.$$

Let $P(x,y):=f(xy+f(xy))=2xf(y)$ $P(0,0)$ yields $f(f(0))=0$ $P(\frac{1}{2},0)$ yields $f(f(0))=f(0)\Longrightarrow f(0)=0$ $P(1,x)$ yields $f(x+f(x))=2f(x)$ $P(x,1)$ yields $f(x+f(x))=2xf(1)$ Therefore from $P(1,x)\text{ and }P(x,1)$ we obtain $2f(x)=2xf(1)\Longrightarrow f(x)=xf(1)\Longleftrightarrow f(x)=cx$ Furthermore by plugging into the original equation we obtain that $c=1\text{ or }c=0$ So, to sum up $\boxed{f(x)=x\text{ and }f(x)=0, \forall x\in\mathbb{R}}$ $\blacksquare$.

parmenides51 wrote: Find all functions $f : R \to R$ that satisfy $f (xy + f(xy)) = 2x f(y)$ for all $x, y \in R$ $\color{blue}\boxed{\textbf{Answer: }f\equiv x \textbf{ or }0}$ $\color{blue}\boxed{\textbf{Proof:}}$ $\color{blue}\rule{24cm}{0.3pt}$ $f(xy+f(xy))=2xf(y)...(\alpha)$ In $(\alpha) y\to x, x\to y:$ $$\Rightarrow f(xy+f(xy))=2yf(x)$$In $(\alpha):$ $$\Rightarrow 2xf(y)=2yf(x)...(\beta)$$In $(\beta) y\to 1:$ $$\Rightarrow 2xf(1)=2f(x)$$$$\Rightarrow f(x)=cx$$Replacing in $(\alpha):$ $$\Rightarrow xyc+xyc^2=2xyc$$$$\Rightarrow c=0\text{ or }1$$$$\Rightarrow \boxed{f\equiv x\text{ or }0}_\blacksquare$$$\color{blue}\rule{24cm}{0.3pt}$