Answer
No, f(x) is not one-to-one.
Work Step by Step
The function $f(x)=\frac{1}{x^2}$ is not one-to-one because it fails the horizontal line test (even function). We can show that it has the same $y$ value for two different $x$ values:
$f(-1)=\frac{1}{(-1)^2}=\frac{1}{1}=1$
$f(1)=\frac{1}{(1)^2}=\frac{1}{1}=1$
Thus the function is not one-to-one.
