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

$ydx+x{y}^{2}dx=xdy$

$ydx-xdy=-x{y}^{2}dx$

$\begin{array}{l}\int \left(\frac{\mathrm{ydx}-\mathrm{xdy}}{{\mathrm{y}}^2}\right)=-\int \mathrm{xdx}\\ \to \int \mathrm{d}(\mathrm{x}/\mathrm{y})=\frac{-{\mathrm{x}}^2}{2}+\mathrm{c}\\ \to \frac{\mathrm{x}}{\mathrm{y}}+\frac{{\mathrm{x}}^2}{2}=\mathrm{c}\text{ which passes through }(1,-1)\\ \to \mathrm{c}=-\frac{1}{2}\\ \Rightarrow \frac{\mathrm{x}}{\mathrm{y}}+\frac{{\mathrm{x}}^2}{2}=-\frac{1}{2}\\ \Rightarrow \text{ put }\mathrm{x}=-\frac{1}{2}\end{array}$

$\begin{array}{l}\Rightarrow \frac{-1}{2y}+\frac{1/4}{2}=-\frac{1}{2}\\ \Rightarrow \frac{-1}{2y}=-\frac{1}{8}-\frac{1}{2}=-\frac{5}{8}\\ \frac{-1}{2y}=-\frac{5}{8}\\ \Rightarrow y=\frac{4}{5}\end{array}$