show that y=ln(ln2x) is a solution of y''+(y')^2 + 1/x(y')=0

This is what I have so far.

y= ln(ln2x)

y'= 1/(xln2x)

y''= 2x/ xln2x

0= 2x/ xln2x + (1/(xln2x))^2 + (1/x)(1/(xln2x))

0= 2x/ xln2x + 1/(xln2x)^2 + 1/ (x^2ln2x)

Then i assume i put the fractions together to get something like

0= (2x(xln2x)(x^2ln2x))/ (xln2x)^2(x^2ln2x))

I know there are more steps which ive tried but from there it gets

very messy.

Am i even on the right track or have i completely missed something?

Thanks.