What is the domain of (sec(x)-1)/(x^2)?

The domain of a function

f(x)=g(x)/h(x) is given by
dom(g(x))∩dom(h(x))\{x|h(x)=0}

which means to find the domain of f(x), we calculate the intersection of the domains of g(x) and h(x), less the points at which h(x)=0 (vertical asymptotes).

In your case,
dom(sec(x)-1) is ℝ,
dom(x²) is also ℝ
All you need to do is to find
dom(sec(x)-1) ∩ dom(x²) but exclude points at which x²=0.