13. Given the function $f(x)$ satisfies $f\left(x+\frac{1}{x}\right)=x^{2}+\frac{1}{x^{2}}$, then the range of $f(x)$ is
Answer $[2,+\infty)$. 
Analysis Let the function $y=f(x)$ satisfy $f\left(t+\frac{1}{t}\right)=t^{2}+\frac{1}{t^{2}}$, then we have

that is,
$$
y=t^{2}+\frac{1}{t^{2}}=\left(t+\frac{1}{t}\right)^{2}-2=x^{2}-2
$$

Therefore, the required function is $f(x)=x^{2}-2(|x| \geq 2)$, and its graph is shown below. It is easy to see that the range of $f(x)=x^{2}-2(|x| \geq 2)$ is $[2,+\infty)$.