I'm assuming I need to use the Method of Dominant Balance to solve this question. Using the substitution $y = e^{S(x)}$, I get $$x^3(e^S S'' + (S')^2e^S) + e^S = x^{-4}$$
In the past, the RHS = 0 and I'd be able to cancel out the $e^S$ terms and go on my way. I'm not sure what to do here and my textbook doesn't have any examples to follow.
For this type of problems, first find the local behavior of the general solution of $$x^3y''+y=0$$ Then you need to add the particular solution. Since the right hand side is the form $x^{-m}$, the solution will be a sum of terms $$y_p=\sum_nc_nx^{-n}$$ Taking the second derivative yields $$y''_p=\sum_n n(n+1)x^{-(n+2)}$$ Plug this in the differential equation and you get $$\sum_nc_n\left(n(n+1)x^{-n+1}+x^{-n}\right)=x^{-4}$$ It is easy to see that the lowest power term is $n=4$. $$y_p=c_0+c_1x^{-1}+c_2x^{-2}+c_3x^{-3}+c_4x^{-4}$$ Identify the coefficients, see if the homogeneous solution has relevant powers, and you have your answer.