Let $P(x,y)$ be assertion. $P(x,-1): f(1)+xf(-x)=f(x)x$ (*) $P(1,x): f(x)+f(-x)=f(1)$ (**) From (*) and (**) we get $f(x)=\frac{c(x+1)}{2x}$

61plus wrote: Find all functions $f:\mathbb{R}\backslash\{0\}\rightarrow\mathbb{R}$ where: $xf(xy)+f(-y)=xf(x)$ for all real $x$ and $y$ not equal to $0$. Let $P(x,y)$ be assertion of $xf(xy)+f(-y)=xf(x)$. $P(1,x)$ $\implies$ $f(x)+f(-x)=f(1)$ $(*)$ $P(x,1)$ \implies $f(-1)=0$ $P(x,-1)$ $\implies$ $xf(-x)+f(1)=xf(x)$ thus $f(x)-f(-x)= \frac{f(1)}{x}$ $(**)$ Summing $(*)$ and $(**)$ and dividing by $2$ yields $f(x)=\frac{f(1)}{2}+\frac{f(1)}{2x}$ so putting $f(1)=c$ we have $f(x)=\frac{c}{2}+\frac{c}{2x}$ which is a solution. So only solution is $f(x)=\frac{c}{2}+\frac{c}{2x}$ where $c$ is real number.