62nd 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: $$\begin{array}{rl}x\cdot f(xy)+f(-y)& =x\cdot c\left(1+\frac{1}{xy}\right)+c\left(1+\frac{1}{-y}\right)=c\left(x+\frac{1}{y}+1-\frac{1}{y}\right)=\\ & =c(x+1)=cx\left(1+\frac{1}{x}\right)=x\cdot f(x).\end{array}$$

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Alternative solution.

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.