Let [tex]y=\displaystyle\sum_{k\ge0}a_kx^k[/tex], so that
[tex]y'=\displaystyle\sum_{k\ge1}ka_kx^{k-1}[/tex]
[tex]y''=\displaystyle\sum_{k\ge2}k(k-1)a_kx^{k-2}[/tex]
Substituting into the ODE gives
[tex]\displaystyle3x\sum_{k\ge2}k(k-1)a_kx^{k-2}+6\sum_{k\ge1}ka_kx^{k-1}+\sum_{k\ge0}a_kx^k=0[/tex]
[tex]\displaystyle3\sum_{k\ge2}k(k-1)a_kx^{k-1}+6\sum_{k\ge1}ka_kx^{k-1}+\sum_{k\ge0}a_kx^k=0[/tex]
The first series starts with a linear term, while the other two start with a constant. Extract the first term from each of the latter two series:
[tex]\displaystyle6\sum_{k\ge1}ka_kx^{k-1}=6\sum_{k\ge2}ka_kx^{k-1}+6a_1[/tex]
[tex]\displaystyle\sum_{k\ge0}a_kx^k=\sum_{k\ge1}a_kx^k+a_0[/tex]
Finally, to get the series to start at the same index, shift the index of the first two series by replacing [tex]k[/tex] with [tex]k+1[/tex]. Then the ODE becomes
[tex]\displaystyle3\sum_{k\ge1}k(k+1)a_{k+1}x^k+6\sum_{k\ge1}(k+1)a_{k+1}x^k+\sum_{k\ge1}a_kx^k+6a_1+a_0=0[/tex]
which can be consolidated to get
[tex]\displaystyle\sum_{k\ge1}\bigg[(3k(k+1)+6(k+1))a_{k+1}+a_k\bigg]x^k+6a_1+a_0=0[/tex]
[tex]\displaystyle\sum_{k\ge1}\bigg[3(k+1)(k+2)a_{k+1}+a_k\bigg]x^k+6a_1+a_0=0[/tex]
You're fixing the solution so that it contains the origin, which means
[tex]y(0)=\displaystyle\sum_{k\ge0}a_kx^k=a_0=0[/tex]
which in turn means [tex]a_1=0[/tex]. With the given recurrence, it follows that [tex]a_k=0[/tex] for all [tex]k\ge2[/tex], so the solution would be [tex]y=0[/tex]. This is to be expected, since [tex]x=0[/tex] is clearly a singular point for the ODE.
