 Solution

Seek a Frobenius solution $y=z^r{\sum }_{n\ge 0}{a}_n{z}^n$ with $a_0\ne 0$. Since $(z+z^3{)}^2=z^2+2z^4+z^6$, its lowest coefficient gives the indicial equationThus the repeated indicial root is $\boxed{r=-1}$. Equating the coefficient of $z^{n+r}$ before setting $r=-1$ givesFor $n\ge 1$ this becomes the linear recurrence relationEvery coefficient is fixed by $a_0$, since the denominator never vanishes for $n\ge 1$. In particular $a_1=0$ and then every odd coefficient is zero. Therefore there is $\boxed{\text{exactly one independent pure Frobenius series}}$, not two merely because the indicial root repeats. If "power series" means an ordinary Taylor series with nonnegative integer powers, the only such solution is zero: its coefficient equation starts with $a_0=0$ and inductively forces all coefficients to vanish.

A second independent local solution is a logarithmic solution from a repeated Frobenius exponent. Indeed, Abel's identity gives $W=C{z}^{-3}[1+O(z^2)]$. For the normalized first solution $y_1=z^{-1}[1+O(z^2)]$, reduction of order givesAfter choosing the harmless multiplicative constant in the numerator, its integrand is $1/s+O(s)$, so $y_2=y_1\log z+z^{-1}\times \text{an analytic series}$. The nonzero logarithmic coefficient proves that this second solution is not another pure Frobenius solution.