1. Rewrite the differential equation

Given $2y e^{x/y^2}\,dx + \left(y^2 - 4x e^{x/y^2}\right)dy = 0.$

Since $x=x(y)$, divide by $dy$: $2y e^{x/y^2}\frac{dx}{dy} + y^2 - 4x e^{x/y^2}=0.$

So, $2y e^{x/y^2}\frac{dx}{dy}=4x e^{x/y^2}-y^2.$

This does not look linear directly, so we try a substitution involving $x/y^2$.

2. Use the substitution

Let $u=\frac{x}{y^2} \quad \Rightarrow \quad x=uy^2.$ Then $\frac{dx}{dy}=y^2\frac{du}{dy}+2uy.$

Also, $e^{x/y^2}=e^u.$

Substitute into the equation: $2y e^u\left(y^2\frac{du}{dy}+2uy\right)+y^2-4(uy^2)e^u=0.$

Expand: $2y^3 e^u\frac{du}{dy}+4uy^2 e^u+y^2-4uy^2 e^u=0.$

The middle terms cancel, giving $2y^3 e^u\frac{du}{dy}+y^2=0.$

Divide by $y^2$: $2y e^u\frac{du}{dy}+1=0.$ So, $\frac{du}{dy}=-\frac{1}{2y e^u}=-\frac{e^{-u}}{2y}.$

Multiply by $e^u$: $e^u\frac{du}{dy}=-\frac{1}{2y}.$

But $\frac{d}{dy}(e^u)=e^u\frac{du}{dy},$ so $\frac{d}{dy}(e^u)=-\frac{1}{2y}.$

3. Integrate

Integrating with respect to $y$: $e^u=-\frac{1}{2}\ln y + C.$

Now use the initial condition $x(1)=0$. At $y=1$, $u=\frac{x}{y^2}=0,$ so $e^u=1.$

Hence, $1=-\frac{1}{2}\ln 1 + C = C.$ Thus, $e^u=1-\frac{1}{2}\ln y.$

Therefore, $u=\ln\left(1-\frac{1}{2}\ln y\right).$ Since $x=uy^2$, $x(y)=y^2\ln\left(1-\frac{1}{2}\ln y\right).$

4. Find $x(e)$

At $y=e$, $\ln e=1.$ So, $x(e)=e^2\ln\left(1-\frac{1}{2}\right)=e^2\ln\left(\frac12\right).$

Using $\ln(1/2)=-\ln 2$, $x(e)=-e^2\ln 2.$

5. Match with the options

$x(e)=-e^2\log_e 2.$ This corresponds to Option D.

6. Comparison with stored answer

Stored correct answer: D

Our derived answer: D

So they agree.