If a curve $y=f(x)$ passes through the point $(1,-1)$ and satisfies the differential equation $y(1+x y) d x=x d y$, then $f\left(-\frac{1}{2}\right)$ is equal to

$\frac{4}{5}$

The differential equation is

$y(1+x y) d x=x d y$

$\Rightarrow y d x-x d y=-x y^2 d x $

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

$\Rightarrow d\left(\frac{x}{y}\right)=-x d x$

On integrating, we obtain

$\frac{x}{y}=-\frac{x^2}{2}+C$          ......(i)

It is given that the curve given by (i) passes thought the point $(1,-1)$.

∴  $-1=-\frac{1}{2}+C \Rightarrow C=-\frac{1}{2}$

Putting $C=-\frac{1}{2}$ in (i), we obtain

$y\left(x^2+1\right)+2 x=0$         .....(ii)

Putting $x=-\frac{1}{2}$ in (ii), we obtain $y=\frac{4}{5}$.

Hence, $f\left(-\frac{1}{2}\right)=\frac{4}{5}$