solve $-2y'(t)=y(1/t)$ spot my mistake:
$$\begin{align}
y'(z)&=-\frac1{t^2} y'(1/t)\\
-\frac{2y'(1/t)}{t^2}&=y(1/t)\\
y'(1/t)-\frac{t^2}2 y(1/t)&=0\\
(y(1/t)e^{-t^3/6})'&=0\\
(y(1/t)e^{-t^3/6})&=c\\
y(1/t)&=ce^{t^3/6}\\
\end{align}$$
thanks