To show the measurability of a function say $f(x)$ we only need to show that the set
$$A= \{x : f(x) > y\}$$
In this case
$$A = \{x : 1/x > y\} = \{x : x < 1/y\}$$
If $y = 0$ the set is empty and empty sets are measurable, I'm not sure what happens if $y$ is not zero
It may be helpful to adopt the slightly different notation $$A_y = \{x\in\mathbb{R}\setminus\{0\}:\; 1/x>y\}.$$ The set $A_y$ is clearly open for every fixed $y\in\mathbb{R}$, and hence measurable. Specifically, if $y>0$ the equation defining $A_y$ can be rearranged as you have done: $$A_y = \{x>0:\; x<1/y\},$$ and if $y=0$, as GEdgar has noted, $A_0$ is not at all empty (and still open) $$A_0 = \{x\in\mathbb{R}\setminus\{0\}:\; 1/x>0\} = \{x\in\mathbb{R}:\; x>0\} = \mathbb{R}^+.$$ Finally if $y<0$, $A_y$ is the union of two open intervals. I should also mention that pretty much any set you'll encounter will naturally be measurable, as in order to produce an example of non-measurable set one has to work real hard.