If you wish to avoid substitutions you can integrate directly. Multiply by $y'/y^2$ to get an exact equation (when integrating I denote $y':=\mathrm dy(\xi)/\mathrm d\xi$ where $\xi$ is a dummy variable): \begin{align} \int_0^x\left( \frac{2y'}{y}y''-\frac{y'^3}{y^2}-y'\right) \mathrm d\xi=0,\\\\ \frac{y'^2}{y}-\frac{(-1)^2}{1}-y+1=\frac{y'^2}{y}-y=0, \end{align} and so \begin{align} y'&=\pm y,\\\\ \int_0^x\frac{y'}{y}\mathrm d\xi&=\int_0^x \pm\mathrm d\xi,\\\\ \log y-\log 1=\pm (x-0)&\implies y=e^{\pm x}. \end{align}

If you wish to avoid substitutions you can integrate directly. Multiply by $y'/y^2$ to get an exact equation (when integrating I denote $y':=\mathrm dy(\xi)/\mathrm d\xi$ where $\xi$ is a dummy variable): \begin{align} \int_0^x\left( \frac{2y'}{y}y''-\frac{y'^3}{y^2}-y'\right) \mathrm d\xi=0,\\\\ \frac{y'^2}{y}-\frac{(-1)^2}{1}-y+1=\frac{y'^2}{y}-y=0, \end{align} and so \begin{align} y'&=-y,\\\\ \int_0^x\frac{y'}{y}\mathrm d\xi&=-\int_0^x \mathrm d\xi,\\\\ \log y-\log 1=- (x-0)&\implies y=e^{- x}. \end{align}