3. (7 points) The function $f$ is such that for any $x>0, y>0$ the equality $f(x y)=f(x)+f(y)$ holds. Find $f(2019)$, if $f\left(\frac{1}{2019}\right)=1$.
Solution. When $y=1 \quad f(x)=f(x)+f(1), f(1)=0$.

When $x=2019 \quad y=\frac{1}{2019} f(1)=f(2019)+f\left(\frac{1}{2019}\right)$,

$f(2019)=f(1)-f\left(\frac{1}{2019}\right)$

$f(2019)=-f\left(\frac{1}{2019}\right)=-1$.

Answer. $f(2019)=-1$.