If the curve y=y(x) represented by the solution of

Given,

$\left(2 x y^{2}-y\right) d x+x d x=0$

$\Rightarrow \frac{d y}{d x}+2 y^{2}-\frac{y}{x}=0$

$\Rightarrow-\frac{1}{y^{2}} \frac{d y}{d x}+\frac{1}{y}\left(\frac{1}{x}\right)=2$

$\frac{1}{y}=z$

$-\frac{1}{y^{2}} \frac{d y}{d x}=\frac{d z}{d x}$

$\Rightarrow \quad \frac{d z}{d x}+z\left(\frac{1}{x}\right)=2$

$\Rightarrow \quad F_{x}=e^{\int \frac{1}{x} d x}=x$

$\Rightarrow z(x)=\int 2(x) d x=x^{2}+c$

$\Rightarrow \frac{x}{y}=x^{2}+c$

As it passes through $\mathrm{P}(2,1)$

[Point of intersection of $2 x-3 y=1$ and $3 x+2 y=8$ ]

$\therefore \frac{2}{1}=4+c$

$\Rightarrow c=-2$

$\Rightarrow \frac{x}{y}=x^{2}-2$

Put $x=1$

$\frac{1}{y}=1-2=-1$

$\Rightarrow y(1)=-1$

$\Rightarrow|y(1)|=1$