3. Find all functions defined on the set of real numbers and taking real values such that for any $x, y$ the equality $f(x+|y|)=f(|x|)+f(y)$ holds.
Answer: $f(x)=0$ for all real $x$.

Sketch of the solution. Let $x=0, y=0$, we get $f(0)=f(0)+f(0)$, so $f(0)=0$. If $y=0$, then $f(x)=f(|x|)$. Let $x=-|y|, 0=f(0)=f(\mid-$ $|y| \mid)+f(y)=f(y)+f(y)=2 f(y)$. Therefore, $f(y)=0$.