How can I solve $y''=x^{-1/3}y^{3/2}$? I am trying to solve the differential equation. 
$$y''=x^{-1/3}y^{3/2}$$
I tried the solution: $y=Ax^s$
$$\implies y'=Asx^{s-1} \\ \iff y''=As(s-1)x^{s-2}$$
$$\implies As(s-1)x^{s-2}= x^{-1/3}(Ax^s)^{3/2} \\ \iff As(s-1)x^{s-2}=x^{-1/3}A^{3/2}x^{\frac{3s}{2}} \\ \iff A^{5/2}s(s-1)=x^{-1/3}x^{3s/2}x^{2-s} \\ \iff Cs(s-1)=x^{\frac{3s}{2}-\frac{1}{3}+2-s} \\ \iff Cs(s-1)=x^{\frac{1}{6}(9s-2+12-6s)}\\ \iff Cs(s-1)=x^{\frac{1}{6}(3s+10)} \\ \iff \ln(Cs^2-Cs)=\frac{1}{6}(3s+10)\ln(x) \\ \iff \ln(C)+\ln(s^2-s)=\frac{1}{6}(3s+10)\ln(x)$$
I am stuck here and I don't know if this is even the best approach. I even tried to solve it with Mathematica but I got no answer. How do I solve this equation?

This is in the form of the Emden-Fowler equation, $y'' = Ax^ny^m$. For $m\neq1$, the solution is
\begin{equation*}
y = Cx^{\frac{n+2}{1-m}}
\end{equation*}
for a constant $C$ that depends on $A$, $n$, and $m$. In your case, the solution is $y = Cx^{\frac{5/3}{-1/2}} = Cx^{-10/3}$. You would have gotten this constant also by requiring $\frac16(3s+10) = 0$, as mentioned by @Winther. Indeed, $y' = C\frac{-10}3x^{-13/3}$, $y'' = C\frac{-10}3\frac{-13}3x^{-16/3}$, and $x^{-1/3}y^{3/2} = x^{-1/3}A^{3/2}x^{-10/2} = A^{3/2}x^{-16/3}$ and so setting $C = \frac{-3}{10}\cdot\frac{-3}{13}\cdot A^{3/2}$ will give you the desired solution. 
