Given $\frac{dy}{2+y}=\frac{-e^{x}dx}{5+e^x}$

$\Rightarrow \int \frac{dy}{2+y}=-\int \frac{e^{x}dx}{5+e^x}$

$\Rightarrow {\log }_e(2+y)=-{\log }_e\left(5+e^x\right)+{\log }_{e}C$

$\Rightarrow {\log }_e(2+y)={\log }_e\left(\frac{C}{5+e^x}\right)$

$\Rightarrow y=\frac{C}{5+e^x}-2$

$\because y\left(0\right)=1$

$\therefore c=18$

$y=\frac{18}{5+e^x}-2$

$y\left({\log }_{e}13\right)=\frac{18}{5+e^{{\log }_e\left(13\right)}}-2$

$\Rightarrow y\left({\log }_{e}13\right)=\frac{18}{5+13}-2$

$\therefore y\left({\log }_{e}13\right)=-1$