Let $P(x,y)$ be the assertation into $f(xy+f(xy)) = xf(y) + yf(x)$ $P(x,1)\Rightarrow f(f(x)+x)=xf(1)+f(x)$ $P(x,y)\Rightarrow xyf(1)+f(xy)=f(xy+f(xy))=xf(y)+yf(x)\Rightarrow \frac{f(1)}{1}+\frac{f(xy)}{xy}=\frac{f(x)}{x}+\frac{f(y)}{y}$ Let $g(x)=\frac{f(x)}{x}$. Then $g(1)+g(xy)=g(x)+g(y)$. Which gives us that $g(x)=k\ln(x)+c$. So $f(x)=x(k\ln(x)+c)$. If $k<0$ then for $x\rightarrow +\infty$ we get a contradiction. From $P(x,\frac{k}{x})$ we get that $f(k+f(k))=xf(\frac{k}{x})+\frac{k}{x}f(x)>\frac{k}{x}f(x) \Rightarrow $ There exists $c_1$ s.t. $f(x)<c_1x$. However if $k>0$ we get that $k\ln(x)+c<c_1$ which is a contradiction for a large enough $x$. Checking $f(x)=cx$ we get the solution, which works $\boxed{f(x)=x}$

GeorgeRP wrote: Let $P(x,y)$ be the assertation into $f(xy+f(xy)) = xf(y) + yf(x)$ $P(x,1)\Rightarrow f(f(x)+x)=xf(1)+f(x)$ $P(x,y)\Rightarrow xyf(1)+f(xy)=f(xy+f(xy))=xf(y)+yf(x)\Rightarrow \frac{f(1)}{1}+\frac{f(xy)}{xy}=\frac{f(x)}{x}+\frac{f(y)}{y}$ Let $g(x)=\frac{f(x)}{x}$. Then $g(1)+g(xy)=g(x)+g(y)$. Which gives us that $g(x)=k\ln(x)+c$. So $f(x)=x(k\ln(x)+c)$. If $k<0$ then for $x\rightarrow +\infty$ we get a contradiction. From $P(x,\frac{k}{x})$ we get that $f(k+f(k))=xf(\frac{k}{x})+\frac{k}{x}f(x)>\frac{k}{x}f(x) \Rightarrow $ There exists $c_1$ s.t. $f(x)<c_1x$. However if $k>0$ we get that $k\ln(x)+c<c_1$ which is a contradiction for a large enough $x$. Checking $f(x)=cx$ we get the solution, which works $\boxed{f(x)=x}$ @above how did you deduce $g(x)=k\ln(x)+c$?

Acorn-SJ wrote: @above how did you deduce $g(x)=k\ln(x)+c$? We can take $g(x)=h(x)+g(1)$. Then we get $h(xy)=h(x)+h(y)$ . Now if we let $h(a^x)=r(x)$ we have that $r(x+y)=r(x)+r(y)$. From there we get that $r$ is linear ($h$ and consequently r are bounded). So $r(x)=vx\Rightarrow h(x)=v\log_a(x) \Rightarrow g(x)=v\log_a(x)+g(1)=v\log_{e^{ln(a)}}(x)+g(1)=\frac{v}{ln_a}\ln(x)+g(1)$

Let $P(x,y)$ denote original equation. From $P(x,y)$ and $P(xy,1)$ we get $$f(xy)+xyf(1)=xf(y)+yf(x)$$Now let $f(x)=xg(x)$ We get $$g(xy)+g(1)=g(x)+g(y)$$By insertion $y=1/x$ we get $$g(x)<g(x)+g(1/x)=2g(1)$$Because of $g$ being bounded from below and above let $A$ be infimum of $g$ and $B$ be supremum. Let $b$ be such that $g(b)=B-\varepsilon$ $$2(B-\varepsilon) = g(1)+g(b^2) \leq B+g(1)$$$$B-2\varepsilon \leq g(1) \leq B$$and by letting $\varepsilon$ be sufficiently small we get $g(1)=B$ Let $a$ be such that $g(a)=A+\varepsilon$ $$2(A+\varepsilon)=B+g(xy) \geq B+A$$$$A+2\varepsilon \geq B \geq A$$and by letting $\varepsilon$ be sufficiently small we get $A=B$ therefore $g$ is constant. Then $f$ must be linear and only solution that works is $f(x)=x$

stmmniko wrote: Let $P(x,y)$ denote original equation. From $P(x,y)$ and $P(xy,1)$ we get $$f(xy)+xyf(1)=xf(y)+yf(x)$$Now let $f(x)=xg(x)$ We get $$g(xy)+g(1)=g(x)+g(y)$$By insertion $y=1/x$ we get $$g(x)<g(x)+g(1/x)=2g(1)$$Because of $g$ being bounded from below and above let $A$ be infimum of $g$ and $B$ be supremum. Let $b$ be such that $g(b)=B-\varepsilon$ $$2(B-\varepsilon) = g(1)+g(b^2) \leq B+g(1)$$$$B-2\varepsilon \leq g(1) \leq B$$and by letting $\varepsilon$ be sufficiently small we get $g(1)=B$ Let $a$ be such that $g(a)=A+\varepsilon$ $$2(A+\varepsilon)=B+g(xy) \geq B+A$$$$A+2\varepsilon \geq B \geq A$$and by letting $\varepsilon$ be sufficiently small we get $A=B$ therefore $g$ is constant. Then $f$ must be linear and only solution that works is $f(x)=x$ Nice logic and well detailed <3