I have tried for the power series but when suppose the solution is $y(x)=\sum _{n=0}^\infty a_nx^n$ appeared sums dependents of $a_n$, $a_{n+2}$ and $a_{n-1}$ and this is not possible.
assume $y(x)=\sum _{n=0}^\infty a_nx^n$ solution, so
$ y'(x) = \sum_{n=1}^\infty na_nx^{n-1}$ and $y''(x)=\sum_{n=2}^\infty n(n-1)a_nx^{n-2}$
then the equation $(1+x^2)y'' - 6xy=0$ is:
$\sum_{n=2}^\infty n(n-1)a_nx^{n-2}+ \sum_{n=2}^\infty n(n-1)a_nx^{n} - \sum_{n=0}^\infty 6a_nx^{n+1}=0$
We can shift the dummy indices, taking k=n−2 in the first sum, k=n in the second sum and k=n+1 in the last sum:
$\sum_{k=0}^\infty(k+2)(k+1)a_{k+2}x^k+\sum_{k=2}^\infty k(k-1)a_{k}x^k- \sum_{k=1}^\infty6a_{k-1}x^k=0$
taking $k=0$ and $k=1$, we have $2a_{2}=0$ so $a_2=0$ and $ 6a_3x-6a_0x=0$ so $a_3=a_0$. For $k>1$ we have
$(k+2)(k+1)a_{k+2} + k(k-1)a_k - 6a_{k-1}=0$
so
$a_k=\frac{ 6a_{k-1} -(k+2)(k+1)a_{k+2} }{k(k-1)}$
and $a_2=0, a_3=a_0$.
but this recorrence not determines the solution cause $a_k$ depends of $a_{k-1}$ and $a_{k+2}$.