The function is undefined only at #x=0# (as there is a denominator of #x#). Then, to look for asymptotes and holes, we will check what happens as #x->0#, as well as the end behavior of the function, that is, what happens when #x->oo# and when #x->-oo#.
First, looking as #x->0#:
Regardless of what happens to #1/x#, we have #-1 <= cos(1/x)<= 1#.
Thus #-x <= xcos(1/x) <= x#, and so
as #x->0#, #xcos(1/x)->0#.
Then there is no asymptote at #x=0#, but #xcos(1/x)# is undefined at that point, meaning #f(x)# has a hole at #x=0#.
Now, looking at the end behavior, #1/x->0# as #x->+-oo#, meaning
#cos(1/x)->cos(0) = 1#
Thus #xcos(1/x) -> +-oo# as #x->+-oo#
So there are no horizontal asymptotes.
But #cos(1/x)# becomes closer and closer to #1# as #x->+-oo#, meaning #f(x)# gets closer and closer to the line #y=x#. Thus #y=x# is a slant asymptote for #f(x)#.
