7. The domain of the function $f(x)$ is $\mathbf{R}$, for any $x \in \mathbf{R}$ and $y \neq 0, f(x+y)=f\left(x y-\frac{x}{y}\right)$, and $f(x)$ is a periodic function, then one of its positive periods is $\qquad$ .
7. $\frac{1+\sqrt{5}}{2} . f(x)$ is a periodic function $\Leftrightarrow$ there must exist $T \neq 0$, such that for any $x \in \mathbf{R}$, $f(x+T)=f(x)$. On the other hand, it is known that $f(x+T)=f\left(x T-\frac{T}{T}\right)=f\left[x\left(T-\frac{1}{T}\right)\right]$. Then $f\left[x\left(T-\frac{1}{T}\right)\right]=f(x)$. As long as $T-\frac{1}{T}=1 \Rightarrow$ $T^{2}-T-1=0$, solving for $T$ gives $T=\frac{1 \pm \sqrt{5}}{2}$. According to the requirement $T>0$. Therefore, $T=\frac{1+\sqrt{5}}{2}$, which means a positive period of $f(x)$ is $\frac{1+\sqrt{5}}{2}$.