The domain of

#y(x) = frac (6x-1) (3x-1)#

is all #RR# except the point #x=1/3# where the denominator vanishes.

Look at the behavior of the function at the limits of the intervals of definition:

#lim_(x->-oo) y(x) = 2#

#lim_(x->+oo) y(x) = 2#

The limits are finite, so #y(x)# has an horizontal asymptote #y=2# on both sides.

#lim_(x->(1/3)^-) y(x) = -oo#

#lim_(x->(1/3)^+) y(x) = +oo#

There is a vertical asymptote at #x=1/3#

The intercepts with the axis can be found as:

#y(x=0) = 1#

#0=frac (6x-1) (3x-1)# at #x=1/6#

The derivative is:

#y'(x)=frac (6(3x-1)-3(6x-1))((3x-1)^2)=-3/(3x-1)^2#

and is always negative, so the function is strictly decreasing in the whole domain.

Summing it up, as #x# goes from #-oo# to #+oo#, #y(x)# starts from below the horizontal line #y=2#, decreases constantly crossing the #y# axis at #y=1# and the #x# axis at #x=1/6#, then goes negative without bound until the vertical line #x=1/3#.

To the right of this line, #y(x)# starts from #+oo# and decreases approaching to the horizontal line #y=2# as #x# grows.