Given differential equation is
$y d x-\left(x+2 y^{2}\right) d y=0$...(1)
and solution of $(1)$ is $x=f(v)$; where $f(-1)=1, f(1)=?$
Rearranging (1), we get
$y \frac{d x}{d y}-\left(x+2 y^{2}\right)=0$
$\Rightarrow \frac{d x}{d y}-2 y-\frac{x}{y}=0$
or $\frac{d x}{d y}+\left(\frac{-1}{y}\right) x=2 y$,
which is a linear differential equation of first order $\frac{d x}{d y}+P$
$x=Q$; Its I.F. $ =e^{\int P d y}=e^{\int \frac{-1}{y} d y}=e^{-\ln y}=\frac{1}{y}$
$\therefore $ Solution of (1) is given by
$x .( I . F )=\int Q( I . F .) dy +C$
$\Rightarrow x \cdot \frac{1}{y}=\int 2 y \cdot \frac{1}{y} dy +C$
$\Rightarrow \frac{x}{y}=2 y +c \Rightarrow x=2 y^{2}+c y ; f(-1)=1$
$x+1=2+c(-1) \Rightarrow c=1 \therefore x=2 y^{2}+y=f(y)$
$\Rightarrow f(1)=2+1=3$