15. Let the domain and range of $f(x)$ both be the set of non-negative real numbers, and satisfy the following conditions:
(1) $f(x \cdot f(y)) \cdot f(y)=f(x+y), x, y \geqslant 0$;
(2) $f(2)=0$;
(3) $f(x) \neq 0$ for all $0 \leqslant x<2$.
Find all functions $f(x)$ that satisfy the above conditions.
15. When $x>2$, take $y=2$, then $f(x)=f((x-2)+2)=f((x-2) \cdot f(2)) \cdot f(2)=0$; when $0 \leqslant x<2$, then $f((2-x) \cdot f(x)) f(x)=f(2)=0$. Since $f(x) \neq 0$, then $f((2-x) \cdot f(x))=0, (2-x) f(x) \geqslant 2$, which gives $f(x) \geqslant \frac{2}{2-x}(0<x<2)$. Also, when $0 \leqslant x<2$, $f\left(\frac{2}{f(x)}+x\right)=f\left(\frac{2}{f(x)} \cdot f(x)\right) f(x)=f(2) f(x)=0$, then $\frac{2}{f(x)}+x \geqslant 2$, which gives $f(x) \leqslant \frac{2}{2-x}$, hence $f(x)=\frac{2}{2-x}$. Therefore, $f(x)=\left\{\begin{array}{l}0, x \geqslant 2 \\ \frac{2}{2-x}, 0 \leqslant x<2 .\end{array}\right.$