Answer
$h$ is also one-to-one.
(see proof below)
Work Step by Step
If we take any two distinct values for x such that $x_{1}\neq x_{2},$ since f is one-to-one, it follows that $f(x_{1})\neq f(x_{2}).$
It also follows that $1/f(x_{1})\neq 1/f(x_{2})$ (both ratios are defined because the denominator is never 0), because if they were equal. then $f(x_{1})$ and $f(x_{2})$ would also be equal, and we know that they are not.
Now, from $1/f(x_{1})\neq 1/f(x_{2})$, it follows that $h(x_{1})\neq h(x_{2})$ by definition of $h$.
Thus, from $x_{1}\neq x_{2}$, it follows that $h(x_{1})\neq h(x_{2})$, which means that $h$ is also one-to-one.
