The particular solution of $\cos ydx+(1+2e^{-x})\sin ydy=0$ when $x=0,y=\frac{\pi}{4}$ is:

$e^x+2=3\sqrt{2}\cos y$

$\cos ydx+(1+2e^{-x})\sin ydy=0⇒-\int\frac{\sin y}{\cos y}dy=\int\frac{dx}{(1+2e^{-x})}$

$⇒log(\cos y)=log(e^x+2)+c⇒\frac{\cos y}{(e^x+2)}=k$ At $x=0,y=\frac{\pi}{4}$

$⇒k=\frac{1}{3\sqrt{2}}$

$⇒e^x+2=3\sqrt{2}\cos y$