Given an ODE of the form
[tex]y''(x)+p(x)y'(x)+q(x)y(x)=f(x)[/tex]
a regular singular point [tex]x=c[/tex] is one such that [tex]p(x)[/tex] or [tex]q(x)[/tex] diverge as [tex]x\to c[/tex], but the limits of [tex](x-c)p(x)[/tex] and [tex](x-c)^2q(x)[/tex] as [tex]x\to c[/tex] exist.
We have for [tex]x\neq0[/tex],
[tex]3x^2y''+2xy'+x^2y=0\implies y''+\dfrac2{3x}y'+\dfrac13y=0[/tex]
and as [tex]x\to0[/tex], we have [tex]x\cdot\dfrac2{3x}\to\dfrac23[/tex] and [tex]x^2\cdot\dfrac13\to0[/tex], so indeed [tex]x=0[/tex] is a regular singular point.
We then look for a series solution about the regular singular point [tex]x=0[/tex] of the form
[tex]y=\displaystyle\sum_{n\ge0}a_nx^{n+k}[/tex]
Substituting into the ODE gives
[tex]\displaystyle3x^2\sum_{n\ge0}a_n(n+k)(n+k-1)x^{n+k-2}+2x\sum_{n\ge0}a_n(n+k)x^{n+k-1}+x^2\sum_{n\ge0}a_nx^{n+k}=0[/tex]
[tex]\displaystyle3\sum_{n\ge2}a_n(n+k)(n+k-1)x^{n+k}+3a_1k(k+1)x^{k+1}+3a_0k(k-1)x^k[/tex]
[tex]\displaystyle+2\sum_{n\ge2}a_n(n+k)x^{n+k}+2a_1(k+1)x^{k+1}+2a_0kx^k[/tex]
[tex]\displaystyle+\sum_{n\ge2}a_{n-2}x^{n+k}=0[/tex]
From this we find the indicial equation to be
[tex](3(k-1)+2)ka_0=0\implies k=0,\,k=\dfrac13[/tex]
Taking [tex]k=\dfrac13[/tex], and in the [tex]x^{k+1}[/tex] term above we find [tex]a_1=0[/tex]. So we have
[tex]\begin{cases}a_0=1\\a_1=0\\\\a_n=-\dfrac{a_{n-2}}{n(3n+1)}&\text{for }n\ge2\end{cases}[/tex]
Since [tex]a_1=0[/tex], all coefficients with an odd index will also vanish.
So the first three terms of the series expansion of this solution are
[tex]\displaystyle\sum_{n\ge0}a_nx^{n+1/3}=a_0x^{1/3}+a_2x^{7/3}+a_4x^{13/3}[/tex]
with [tex]a_0=1[/tex], [tex]a_2=-\dfrac1{14}[/tex], and [tex]a_4=\dfrac1{728}[/tex].
