Given
$$
f(x y)=f(x)+f(y)
$$

and $f(2)=1$. Find the value of $f\left(\frac{1}{64}\right)$.
By setting $x=y=1$ in (1), we know that $f(1)=2 f(1), f(1)=0$. By setting $y=\frac{1}{x}$, we know that $f(1)=f(x)+f\left(\frac{1}{x}\right)$, so when $x \neq 0$, we have $f(x)=-f\left(\frac{1}{x}\right)$. Therefore, $f\left(\frac{1}{64}\right)=-f(64)$. By setting $(x, y)=(2,2),(2,4),(8,8)$ in (1) successively, we know that $f(4)=2 f(2)=2$, $f(8)=f(4)+f(2)=3, f(64)=2 f(8)=6$. Therefore, $f\left(\frac{1}{64}\right)=-6$.