Answer
V.A. $x=0$
Work Step by Step
Given $y=\frac{x^4-1}{x^2-x}=\frac{(x^2+1)(x+1)(x-1)}{x(x-1)}=\frac{(x^2+1)(x+1)}{x}, x\ne1$,
we can find its vertical asymptote(s) V.A. $x=0$, horizontal asymptote(s) H.A. $none$, oblique asymptote(s) O.A. $none$.