jmc

Note that $f(x)=\frac{1}{x^2}>0$ for all nonzero $x$. That is, the range of $f$ must only include positive numbers. Conversely, if $a$ is a positive number, then $$f\left(\frac{1}{\sqrt{a}}\right)=\frac{1}{(1/\sqrt{a}{)}^2}=a,$$so $a$ is indeed in the range of $f$. Thus, the range of $f$ is the set of all positive real numbers; in interval notation, that's $\boxed{(0,\infty )}$.