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