There are two vertical asymptotes (#x=-3# and #x=3#) and one horizontal asymptote (#y=0#).

We have #f(x) = (x^2-1)/(x^4-81)=((x+1)(x-1))/((x^2+9)(x+3)(x-3))#.

Therefore, the domain of #f(x)# is :

#D = ]-oo;-3[ uu ]-3;3[ uu ]3;+oo[#.

In order to find the **vertical asymptotes**, you need to examine

#lim_(x->a)|f(x)|#, where #a# is a forbidden #x#-value of #f(x)#.

(In general, you'll take forbidden #x#-values from the edges of the domain, excluding #+oo# and #-oo#).

Here, the forbidden values at the edges of the domain are #a_1 = -3# and #a_2 = 3#.

There is a vertical asymptote in #x=a# if #lim_(x->a)|f(x)|=+oo#.

#lim_(x->-3)|f(x)|=((-2)*(-4))/(18*0*(-6))=8/0 = +oo#.

#lim_(x->+3)|f(x)|=(4*2)/(18*6*0)=8/0 = +oo#.

Therefore, there are two vertical asymptotes,

one in #x = -3# and another in #x = 3#.

In order to find the **horizontal asymptotes**, you need to examine

#lim_(x->-oo)f(x)# and #lim_(x->+oo)f(x)#.

There is an horizontal asymptote in #y=h# on the left of your graph/function if #lim_(x->-oo)f(x)=h#.

#lim_(x->-oo)f(x)=lim_(x->-oo)(x^2-1)/(x^4-81) = lim_(x->-oo)x^2/x^4 = lim_(x->-oo)1/x^2 = 1/(+oo) = 0#.

And there is an horizontal asymptote in #y=h# on the right of your graph/function if #lim_(x->+oo)f(x)=h#.

#lim_(x->+oo)f(x)=lim_(x->+oo)(x^2-1)/(x^4-81) = lim_(x->+oo)x^2/x^4 = lim_(x->+oo)1/x^2 = 1/(+oo) = 0#.

Therefore, there is one horizontal asymptote when #y=0#, on both sides of the function.

Since there are already horizontal asymptotes on the left and on the right of the function, there won't be any **oblique asymptotes** on both sides of it.

That's it.