Find all functions $f(x)$ defined on the set of rational numbers whose function values are rational numbers, satisfying $f(x-f(y))=$ $f(x) \cdot f(y), x, y \in \mathbf{Q}$.
15. Clearly, for all rational numbers $x, f(x)=0$ satisfies the condition.

Assume there exists a rational number $y, f(y) \neq 0$, which implies that for any $y \in \mathbf{Q}$, we have $f(y) \neq 0$. In this case, let $x = -f(0), y=0$, and we can find that there exists $a \in \mathbf{Q}, f(a)=1$. From this, we can deduce that $f(x+1)=f(x), x \in \mathbf{Q}$.
Thus, $f(x)$ is a periodic function with a period of 1, and we have $f(n f(y))=\cdots=f(0) f^{n}(y)$.
Let $f(y)=\frac{q}{p}, p>0, p, q \in \mathbf{Z}$, then $f^{\circ}(y)=1$.
From this, we can infer that for any rational number $x$, we have $f(x)=1$.
Therefore, the functions that satisfy the condition are only two: $f(x)=0$ and $f(x)=1$.