Solving the DE: $y''+1/(2y^3)=0$. \begin{align}
y''+\frac{1}{2y^3}=0\\
\end{align}
This is a second order differential equation which doesn't contain $x$ explicitly.
Let $y'=p(y)$, so that $y''=p(y)*p'(y)$.
\begin{align}
pp'+\frac{1}{2y^3}=0\\
\int{pdp}=\int{-\frac{1}{2y^3}dy}\\
\frac{p^2}{2}=-\frac{1}{2}*\frac{-1y^{-2}}{2}+C_1\\
p^2=\frac{1}{2y^2}+C_1\\
p=\pm\sqrt{\frac{1+C_1y^2}{2y^2}}\\
\end{align}
$p$ was equal to $y'$
\begin{align}
y'=\pm\sqrt{\frac{1+C_1y^2}{2y^2}}\\
\end{align}
I've tried to continue with trigonometric substitution but that didn't work for me.
Can you please help me out?
Thank you in advance!
Edit: adding final solution
Because of the answers below, I found the solution:
\begin{align}
\frac{\sqrt2ydy}{\sqrt{1+C_1y^2}}=\pm{x}dx\\
\int{\frac{\sqrt2ydy}{\sqrt{1+C_1y^2}}}=\int{\pm{x}dx}\\
\frac{1}{2C_1}\int{\frac{d({1+C_1y^2})}{\sqrt{1+C_1y^2}}}=\pm\frac{1}{\sqrt2}x\\
2\sqrt{1+C_1y^2}=\pm{\frac{2C_1}{\sqrt{2}}}(x+C_2)\\
\sqrt{1+C_1y^2}=\pm{\frac{C_1}{\sqrt{2}}}(x+C_2)\\
2(1+C_1y^2)=C_1^2(x+C_2)\\
\end{align}

Try to find $x$ as a function of $y^2$ instead, for taking it next from there, in terms of $y^2$:
$$ dx= \frac{\sqrt2y dy}{\sqrt{1+2 C y^2}} = \frac{d y^2}{\sqrt{2(1+2 C y^2})} $$
leading to inverse circular/hyperbolic function solutions.
