Given equation is:

$({\mathrm{x}}^2– 3{\mathrm{y}}^2)\mathrm{dx}+3\mathrm{xydy}=0$

We can re-write equation as

$\frac{\mathrm{dy}}{\mathrm{dx}}=-\frac{\left({\mathrm{x}}^2-3{\mathrm{y}}^2\right)}{3\mathrm{xy}}$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{y}}{\mathrm{x}}-\frac{1}{3}\frac{\mathrm{x}}{\mathrm{y}}$  ....(1)

Put $\mathrm{y}=\mathrm{vx}$

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

Equation (1) can be written as

$$\begin{gathered} \mathrm{v}+\mathrm{x}\frac{\mathrm{dv}}{\mathrm{dx}}=\mathrm{v}-\frac{1}{3}\frac{1}{\mathrm{v}} \\ \Rightarrow \mathrm{vdv}=-\frac{1}{3\mathrm{x}} \end{gathered}$$

Integrating both sides, we get

$\frac{{\mathrm{v}}^2}{2}=-\frac{1}{3}\ln \left(\mathrm{x}\right)+\mathrm{c}$

$\Rightarrow \frac{{\mathrm{y}}^2}{2{\mathrm{x}}^2}=\frac{-1}{3}\ln \left(\mathrm{x}\right)+\mathrm{c}$        .....(2)

$\because \mathrm{y}\left(1\right)=1$     (given)

$\therefore \mathrm{c}=\frac{1}{2}$       (from equation 2)

Equation (2) can be written as

$\frac{{\mathrm{y}}^2}{2{\mathrm{x}}^2}=\frac{-1}{3}\ln \left(\mathrm{x}\right)+\frac{1}{2}$

$\Rightarrow {\mathrm{y}}^2=\frac{-2}{3}{\mathrm{x}}^2\ln \left(\mathrm{x}\right)+{\mathrm{x}}^2$

Now ${\mathrm{y}}^2\left(\mathrm{e}\right)=\frac{-2}{3}{\mathrm{e}}^2\ln \left(\mathrm{e}\right)+{\mathrm{e}}^2=\frac{-2}{3}{\mathrm{e}}^2+{\mathrm{e}}^2=\frac{{\mathrm{e}}^2}{3}$

$\Rightarrow 6{\mathrm{y}}^2\left(\mathrm{e}\right)=2{\mathrm{e}}^2$