Find all functions $f: \mathbb{R} \backslash\{0\} \rightarrow \mathbb{R}$ for which:

$$
x f(x y)+f(-y)=x f(x)
$$

for all real $x, y$ not equal to 0.
Substitute $x=1$, which gives $f(y)+f(-y)=f(1)$, or $f(-y)=f(1)-f(y)$ for all $y$. Now substitute $y=-1$, which gives $x f(-x)+f(1)=x f(x)$. If we substitute $f(-x)=$ $f(1)-f(x)$ here, we get $x(f(1)-f(x))+f(1)=x f(x)$, so $x f(1)+f(1)=2 x f(x)$. We see that $f$ is of the form $f(x)=c+\frac{c}{x}$ for some $c \in \mathbb{R}$. We check this family of functions. On the left, we have $x f(x y)+f(-y)=x\left(c+\frac{c}{x y}\right)+c+\frac{c}{-y}=x c+c$ and on the right, we have $x f(x)=x\left(c+\frac{c}{x}\right)=x c+c$. These two are equal, so this function satisfies for all $c \in \mathbb{R}$.