Czech and Slovak Mathematical Olympiad

Substituting $x = 1$ gives
$$f(y) + f(-y) = f(1).$$
Denoting $f(1) = a$, we have $f(-y) = a - f(y)$. Substituting $y = -1$, we get
$$x \cdot f(-x) + f(1) = x \cdot f(x),$$
i. e.,
$$x(a - f(x)) + a = x \cdot f(x),$$
hence
$$f(x) = \frac{a(x+1)}{2x} = \frac{a}{2}\left(1+\frac{1}{x}\right).$$
Finally, we check that for any real number $c$, the function $f(x) = c(1+1/x)$ satisfies the conditions:
$$\begin{aligned} x \cdot f(xy) + f(-y) &= x \cdot c\left(1 + \frac{1}{xy}\right) + c\left(1 + \frac{1}{-y}\right) = c\left(x + \frac{1}{y} + 1 - \frac{1}{y}\right) = \\ &= c(x+1) = cx\left(1 + \frac{1}{x}\right) = x \cdot f(x). \end{aligned}$$

Let us set $f(1) = a$. Substituting $y = -1$ into the given equation yields
$$xf(-x) + a = xf(x),$$
i. e.,
$$f(-x) = f(x) - \frac{a}{x}.$$

The given equation can be rearranged into the form
$$f(xy) + \frac{1}{x} \cdot f(-y) = f(x),$$
whence, using (1), we have
$$f(xy) + \frac{1}{x} \left( f(y) - \frac{a}{y} \right) = f(x),$$
$$f(xy) + \frac{f(y)}{x} - \frac{a}{xy} = f(x).$$

Interchanging $x$ and $y$ gives
$$f(yx) + \frac{f(x)}{y} - \frac{a}{yx} = f(y),$$
so, combining the last two equations, we get
$$\frac{f(x)}{y} - \frac{f(y)}{x} = f(y) - f(x),$$
and substituting $y = 1$ now gives
$$2f(x) = a\left(1 + \frac{1}{x}\right).$$
Again, we can easily verify that every function $f(x) = c(1 + 1/x)$ is a solution of the given functional equation.