Question

find two linearly independent solutions to the differential equation (3x-1)^2y"+(9x-3)y'-9y=0

Answer 1

Not sure that this is what is required.
From Mathematica v9:

Answer 2

If we change variable, namely:
\[\large 3x - 1 = s\]
where s is a new variable of differentiation, we can rewrite your differential equation as below:
\[\large {s^2}{y_{ss}} + s{y_s} - y = 0\]
where
\[\large {y_s} = \frac{{dy}}{{ds}},\;{y_{ss}} = \frac{{{d^2}y}}{{d{s^2}}}\]
Now, that last form of your equation is a homogeneous Euler equation.
In order to find its solutions, we try this substitution:
\[\large y = k{s^a}\]
After that substitution, we get:
\[\large a = 1,\quad a = - 1\] (please check those values)
so two linearly independent solutions, are:
\[\large y = s,\quad y = \frac{1}{s}\]
or using the definition of s:
\[\large {y_1} = 3x - 1,\quad {y_2} = \frac{1}{{3x - 1}}\]