Hint:
$\displaystyle z=\ln(x)$
Use the chain rule
$\displaystyle \frac{dy}{dx}=\frac{dy}{dz}\frac{dz}{dx}=\frac{1}{ x}\frac{dy}{dz}$
Now we need the 2nd derivative using the product and chain rule
$\displaystyle \frac{d^2y}{dx^2}=-\frac{1}{x^2}\frac{dy}{dz}+\frac{1}{x}\left( \frac{d^2y}{dz^2} \cdot \frac{1}{x}\right)$
from here sub into the equation
Good luck
$\displaystyle y=\sum_{n=0}^{\infty}c_nx^{n+r} $
$\displaystyle y'=\sum_{n=0}^{\infty}c_n(n+r)x^{n+r-1} $
$\displaystyle y''=\sum_{n=2}^{\infty}c_n(n+r)(n+r-1)x^{n+r-2} $
$\displaystyle x^2y''+xy'+y=0$
$\displaystyle x^2\sum_{n=0}^{\infty}c_n(n+r)(n+r-1)x^{n+r-2}+x\sum_{n=0}^{\infty}c_n(n+r)x^{n+r-1}+ \sum_{n=0}^{\infty}c_nx^{n+r}=0$
$\displaystyle \sum_{n=0}^{\infty}c_n(n+r)(n+r-1)x^{n+r}+\sum_{n=0}^{\infty}c_n(n+r)x^{n+r}+ \sum_{n=0}^{\infty}c_nx^{n+r}=0$
$\displaystyle c_nx^{n+r}\left(\sum_{n=0}^{\infty}(n+r)(n+r-1)+(n+r)+ 1 \right)=0 $
$\displaystyle c_nx^{n+r}\left(\sum_{n=0}^{\infty}(n+r)[(n+r-1)+1]+ 1 \right)=0 $
$\displaystyle c_nx^{n+r}\left(\sum_{n=0}^{\infty}(n+r)^2+ 1 \right)=0 $
I think this is where you are stuck.
Thanks Mr. Fantastic,
I forgot that $\displaystyle c_0 \ne 0$ in this method so
$\displaystyle c_nx^{n+r}\left(\sum_{n=0}^{\infty}(n+r)^2+ 1 \right)=0 \iff c_0x^{r}(r^2+1)+c_nx^{n+r}\left(\sum_{n=1}^{\infty }(n+r)^2+ 1 \right)=0$
so $\displaystyle c_n=0, \forall n \ge 1$
$\displaystyle r^2=1 \iff r=\pm i$
Then the rest follows.
Thanks again!