What are the asymptote(s) and hole(s), if any, of  f(x) =(x^2-1)/(x^4-1) ?

Double asymptote $y = 0$
$f \left(x\right) = \frac{{x}^{2} - 1}{{x}^{4} - 1} = \frac{{x}^{2} - 1}{\left({x}^{2} + 1\right) \left({x}^{2} - 1\right)} = \frac{1}{{x}^{2} + 1}$
So $f \left(x\right)$ has a double asymptote characterized as $y = 0$