4. Find all functions $f: \mathbb{R} \backslash\{0\} \rightarrow \mathbb{R}$ such that for all non-zero numbers $x, y$ the following holds:

$$
x \cdot f(x y)+f(-y)=x \cdot f(x) .
$$
SOLUTION. By substituting $x=1$ we get

$$
f(y)+f(-y)=f(1) .
$$

If we denote $f(1)=a$, we have $f(-y)=a-f(y)$. Substituting $y=-1$ as well, we get

$$
x \cdot f(-x)+f(1)=x \cdot f(x)
$$

which means

$$
x(a-f(x))+a=x \cdot f(x)
$$

and from this

$$
f(x)=\frac{a(x+1)}{2 x}=\frac{a}{2}\left(1+\frac{1}{x}\right) .
$$

By verification, we find that for any real $c$, every function $f(x)=c(1+1 / x)$ satisfies:

$$
\begin{aligned}
x \cdot f(x y)+f(-y) & =x \cdot c\left(1+\frac{1}{x y}\right)+c\left(1+\frac{1}{-y}\right)=c\left(x+\frac{1}{y}+1-\frac{1}{y}\right)= \\
& =c(x+1)=c x\left(1+\frac{1}{x}\right)=x \cdot f(x) .
\end{aligned}
$$