$$\begin{aligned} & \left(1+y^2\right) e^{\tan x} d x+\cos ^2 x\left(1+e^{2 \tan x}\right) d y=0 \\ & \frac{d y}{1+y^2}=-\frac{e^{\tan x} \cdot \sec ^2 x d x}{1+e^{2 \tan x}} \\ & \int \frac{d y}{1+y^2}=-\int \frac{e^{\tan x} \cdot \sec ^2 x d x}{1+e^{2 \tan x}} \end{aligned}$$

Let $$e^{\tan x}=t$$

$$\begin{aligned} & e^{\tan x} \cdot \sec ^2 x d x=d t \\ & \int \frac{d y}{1+y^2}=-\int \frac{d t}{1+t^2} \end{aligned}$$

$$\begin{aligned} & \tan ^{-1} y=-\tan ^{-1} t+c \\ & \tan ^{-1} y=-\tan ^{-1}\left(e^{\tan x}\right)+c \\ & \text { at } x=0, y=1 \\ & \tan ^{-1}(1)=-\tan ^{-1}(1)+c \\ & \frac{\pi}{4}=-\frac{\pi}{4}+c \\ & c=\frac{\pi}{2} \\ & \tan ^{-1} y=-\tan ^{-1}\left(e^{\tan x}\right)+\frac{\pi}{2} \end{aligned}$$

Now, at $$x=\frac{\pi}{4}$$

$$\begin{aligned} & \tan ^{-1} y=-\tan ^{-1}(e)+\frac{\pi}{2} \\ & \tan ^{-1} y=\cot ^{-1} e=\tan ^{-1} \frac{1}{e} \\ & \Rightarrow y=\frac{1}{e} \end{aligned}$$