7. Given that the function $f(x)$ is strictly monotonically decreasing on $(0,+\infty)$, for any $x \in(0,+\infty)$, we have $f(x) \cdot f\left(f(x)+\frac{2}{x}\right)=\frac{1}{3}$. Let $g(x)=f(x)+4 x^{2}, x \in(0,+\infty)$, then the minimum value of the function $g(x)$ is
Let $t=f(x)+\frac{2}{x}$, according to the problem, we have $f(x)=\frac{1}{3 f(t)}$. Substituting $x=t$ into the original equation, we get $f(t) \cdot f\left(f(t)+\frac{2}{t}\right)=\frac{1}{3}$ $\Rightarrow f\left(f(t)+\frac{2}{t}\right)=\frac{1}{3 f(t)}=f(x)$. Since the function $f(x)$ is strictly decreasing on $(0,+\infty)$, we have $f(t)+\frac{2}{t}=x \Rightarrow \frac{1}{3 f(x)}+\frac{2}{f(x)+\frac{2}{x}}=x \Rightarrow f(x)+\frac{2}{x}+6 f(x)=3 x f(x)\left(f(x)+\frac{2}{x}\right)$ $\Rightarrow 3 x f^{2}(x)-f(x)-\frac{2}{x}=0 \Rightarrow(x f(x)-1)\left(3 f(x)+\frac{2}{x}\right)=0 \Rightarrow f(x)=\frac{1}{x}$ or $f(x)=-\frac{2}{3 x}$ (discard). Therefore, $g(x)=\frac{1}{x}+4 x^{2}=\frac{1}{2 x}+\frac{1}{2 x}+4 x^{2} \geqslant 3$, equality holds when $x=\frac{1}{2}$.