= Solution
Summing the <geometric series> gives
$$
\boxed{y(x)=\frac1{1-x^3}.}
$$
There is a particularly short verification using the <product rule>:
$$
(1-x^3)y''-6x^2y'-6xy=\bigl[(1-x^3)y\bigr]''=1''=0.
$$
Also $y(0)=1$ and $y'(0)=0$. The <radius of convergence> of the series is one, set by the three complex cube roots of unity; the closed expression continues the solution beyond that disk wherever its denominator is nonzero. The real initial-value solution extends on the interval $(-\infty,1)$.
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