$\frac{d y}{d x}+\frac{y}{x}=\frac{x^2}{x^2}+\frac{y^2}{x^2}$

$\begin{aligned} & \frac{d y}{d x}+\frac{y}{x}=1+\left(\frac{y}{x}\right)^2 \\ & \text { Let } \frac{y}{x}=t \\ & y=x t \\ & \frac{d y}{d x}=x \frac{d t}{d x}+t \\ & \therefore x \frac{d t}{d x}+t+t=1+t^2 \\ & \Rightarrow x \frac{d t}{d x}+2 t=1+t^2 \\ & \Rightarrow x \frac{d t}{d x}+t^2+1-2 t \\ & \Rightarrow x \frac{d t}{d x}=(t-1)^2 \\ & \Rightarrow \frac{d t}{(t-1)^2}=\frac{d x}{x}\end{aligned}$

Integrating both sides

$$ \begin{aligned} & \Rightarrow \int \frac{d t}{(t-1)^2}=\int \frac{d x}{x} \\ & \Rightarrow \frac{-1}{(t-1)}=\ln x+C \\ & \Rightarrow \frac{-1}{\frac{y}{x}-1}=\ln x+C \\ & \Rightarrow \frac{-x}{y-x}=\ln x+C \end{aligned} $$

Given $y(1)=0$

$$ \begin{aligned} & 1=C \\ & \Rightarrow \frac{-x}{y-x}=\ln x+1 ..........(i)\end{aligned} $$

$\therefore $ Put $x=e$

$\begin{aligned} & \frac{-e}{y-e}=1+1 \\ & \Rightarrow-e=2(y-e) \\ & \Rightarrow e=2(e-y) \\ & \Rightarrow \frac{e}{2}=e-y \\ & \Rightarrow y=e-\frac{e}{2} \Rightarrow \frac{e}{2} \Rightarrow y=\frac{e}{2} \\ & \text { Put } x=e^2 \text { in (i) } \\ & \Rightarrow \frac{-e^2}{y-e^2}=2+1\end{aligned}$

$\begin{aligned} & \Rightarrow-e^2=3\left(y-e^2\right) \\ & \Rightarrow-e^2=3 y-3 e^2 \\ & \Rightarrow 2 e^2=3 y \\ & \Rightarrow y=\frac{2}{3} e^2 \\ & \therefore \frac{2(y(e))^2}{y\left(e^2\right)}=2 \frac{\left(\frac{e}{2}\right)^2}{\frac{2}{3} e^2} \\ & \Rightarrow \frac{\frac{1}{2}}{\frac{2}{3}}=\frac{3}{4}=00.75\end{aligned}$