= Solution
After normalization, $p=-6x^2/(1-x^3)$ and $q=-6x/(1-x^3)$. Both are analytic at zero, so $\boxed{x=0\text{ is ordinary}}$. At one,
$$
(x-1)p=\frac{6x^2}{1+x+x^2},\qquad (x-1)^2q=\frac{6x(x-1)}{1+x+x^2}
$$
are analytic, whereas $p,q$ themselves have <poles>. Hence $\boxed{x=1\text{ is a regular singular point}}$.
Insert $y=\sum_{n\geq0}a_nx^n$. The coefficient of $x^0$ gives $a_2=0$; for $k\geq1$ the coefficient of $x^k$ gives
$$
(k+2)(k+1)a_{k+2}=(k+2)(k+1)a_{k-1}.
$$
Thus $a_n=a_{n-3}$ for $n\geq3$. The initial data give $a_0=1$, $a_1=0$, so
$$
\boxed{y(x)=1+x^3+x^6+\cdots=\frac1{1-x^3},\qquad |x|<1.}
$$
The radius is one because the three cube roots of unity are <poles> of this solution. A direct structural check observes that the differential equation is exactly $[(1-x^3)y]''=0$. Thus $(1-x^3)y=A+Bx$, and the initial data select $A=1,B=0$. This is the <product-derivative reduction of a second-order differential equation>.
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