1. Given differential equation

$x^2\frac{dy}{dx}+xy=x^2+y^2, \qquad x>\frac1e, \qquad y(1)=0$

We need to find

$2\frac{(y(e))^2}{y(e^2)}.$

2. Rewrite the equation

Divide by $x^2$:

$\frac{dy}{dx}+\frac{1}{x}y=1+\left(\frac{y}{x}\right)^2.$

This suggests the substitution

$v=\frac{y}{x} \quad \Rightarrow \quad y=vx.$

Then

$\frac{dy}{dx}=v+x\frac{dv}{dx}.$

Substitute into the differential equation:

$v+x\frac{dv}{dx}+v=1+v^2.$

So,

$x\frac{dv}{dx}=v^2-2v+1=(v-1)^2.$

Hence,

$\frac{dv}{(v-1)^2}=\frac{dx}{x}.$

3. Integrate

Integrating both sides,

$\int \frac{dv}{(v-1)^2}=\int \frac{dx}{x}.$

Now,

$\int (v-1)^{-2}dv=-\frac{1}{v-1}.$

Therefore,

$-\frac{1}{v-1}=\ln x + C.$

Use the initial condition $y(1)=0$. At $x=1$,

$v(1)=\frac{y(1)}{1}=0.$

So,

$-\frac{1}{0-1}=\ln 1 + C \Rightarrow 1=0+C \Rightarrow C=1.$

Thus,

$-\frac{1}{v-1}=\ln x+1.$

Rearrange:

$\frac{1}{v-1}=-(\ln x+1)$

$v-1=-\frac{1}{\ln x+1}$

$v=1-\frac{1}{\ln x+1}=\frac{\ln x}{\ln x+1}.$

Since $v=\dfrac{y}{x}$,

$y=x\cdot \frac{\ln x}{\ln x+1}.$

4. Compute $y(e)$ and $y(e^2)$

For $x=e$,

$\ln e=1$

so

$y(e)=e\cdot \frac{1}{2}=\frac{e}{2}.$

For $x=e^2$,

$\ln(e^2)=2$

so

$y(e^2)=e^2\cdot \frac{2}{3}=\frac{2e^2}{3}.$

5. Evaluate the required expression

$2\frac{(y(e))^2}{y(e^2)}=2\cdot \frac{\left(\frac e2\right)^2}{\frac{2e^2}{3}}.$

Now,

$\left(\frac e2\right)^2=\frac{e^2}{4}.$

Hence,

$2\cdot \frac{e^2/4}{2e^2/3}=2\cdot \frac{e^2}{4}\cdot \frac{3}{2e^2}.$

Cancel $e^2$:

$=2\cdot \frac{3}{8}=\frac{3}{4}.$

So the required value is

\boxed{\frac34}}

Since this is an integer-type question with range-style stored answer, numerically

$\frac34=0.75.$

6. Comparison with stored correct answer

Stored correct answer: 0.7TO0.8

Our derived value is $0.75$, which lies in this range. So it agrees.