If $y=y\left(x\right)$ is the solution of the differential equation $2x^2\frac{dy}{dx}-2xy+3y^2=0$ such that $y\left(e\right)=\frac{e}{3}$, then $y\left(1\right)$ is equal to

Given $2x^2\frac{dy}{dx}-2xy+3y^2=0$

Dividing by $2x^2{y}^2$, we get,

$-\frac{1}{y^2}\frac{dy}{dx}+\frac{1}{xy}=\frac{3}{2x^2}$

Let $\frac{1}{y}=p,\frac{-1}{y^2}\frac{dy}{dx}=\frac{dp}{dx}$

$\Rightarrow \frac{dp}{dx}+\frac{1}{x}\left(p\right)=\frac{3}{2x^2}$

General solution will be $p·e^{\int \frac{1}{x}dx}=\int \frac{3}{2x^2}{·}^{e^{\int \frac{1}{x}dx}}dx+c$
$\Rightarrow \frac{x}{y}=\int \frac{3}{2x}dx+c \Rightarrow \frac{x}{y}=\frac{3}{2} \ln x+c$

$\because y\left(e\right)=\frac{e}{3}\Rightarrow c=\frac{3}{2} \Rightarrow \frac{x}{y}=\frac{3}{2}\ln x+\frac{3}{2}$ is the particular solution.

when $x=1; y\left(1\right)=\frac{2}{3}$