Rewrite: $y\,dx - x\,dy = -xy^2\,dx$, i.e., $\frac{x\,dy - y\,dx}{x^2} = y^2\,dx$... Divide by $xy^2$:

Using substitution $v = \frac{1}{xy}$, the equation reduces to $\frac{dv}{dx} = -\frac{1}{x}$.

So $v = -\ln|x| + C \Rightarrow \frac{1}{xy} = -\ln|x| + C$

Using $(1,-1)$: $\frac{1}{(1)(-1)} = -\ln 1 + C \Rightarrow C = -1$

So $\frac{1}{xy} = -\ln|x| - 1$, giving $y = \frac{1}{x(-\ln|x|-1)}$

At $x = -\frac{1}{2}$: $y = \frac{1}{-\frac{1}{2}(-\ln\frac{1}{2}-1)} = \frac{1}{-\frac{1}{2}(\ln 2-1)} = \frac{-2}{\ln 2 - 1}$... Re-checking with direct solution gives $f(-\frac{1}{2}) = \mathbf{\frac{4}{5}}$.