$$ \begin{aligned} & \left(2 x y^2-y\right) d x+x d y=0 \\ & 2 x y^2 d x-y d x+x d y=0 \\ & 2 x d x=\frac{y d x-x d y}{y^2} d\left(\frac{x}{y}\right) \end{aligned} $$ Now integrate $$ x^2=\frac{x}{y}+c $$ Now point of intersection of lines are $(2,1)$ $$ \begin{aligned} & 4=\frac{2}{1}+c \quad \Rightarrow c=2 \\ & x^2=\frac{x}{y}+2 \end{aligned} $$ Now $y(1)=-1$ $$ \Rightarrow \quad|y(1)|=1 $$