Answer
$x=0, -1 $ and $1$ are vertical asymptotes.
$y=-1$ is horizontal asymptotes.
Work Step by Step
\[y=\frac{1+x^4}{x^2-x^4}\]
For vertical asymptotes denominator should be zero.
\[x^2-x^4=0\]
\[\Rightarrow x^2(1-x^2)=0\]
\[\Rightarrow x=0,-1 \;\text{and} \;1\]
Therefore $x=0, x=-1$ and $x=1$ are vertical asymototes.
Let \[L=\lim_{x\rightarrow\infty}\displaystyle\frac{1+x^4}{x^2-x^4}\]
\[L=\lim_{x\rightarrow\infty}\frac{\displaystyle\frac{1+x^4}{x^4}}{\displaystyle\frac{x^2-x^4}{x^4}}\]
\[L=\lim_{x\rightarrow\infty}\displaystyle\frac{\displaystyle\frac{1}{x^4}+1}{\displaystyle\frac{1}{x^2}-1}\]
\[L=\frac{0+1}{0-1}=-1\]
Therefore $y=-1$ is horizontal asymptote.