$x \mathrm{~d} y=y(\mathrm{~d} x+y \mathrm{~d} y)$
$\Rightarrow y \mathrm{~d} x=\left(x-y^2\right) \mathrm{d} y \Rightarrow \frac{\mathrm{d} x}{\mathrm{~d} y}+\left(-\frac{1}{y}\right) x=-y$
$\therefore \quad$ I.F. $=\mathrm{e}^{\int-\frac{1}{d y} y^y}=\mathrm{e}^{-\log y}=\frac{1}{y}$
$\therefore \quad$ Solution of the given equation is
$x \cdot \frac{1}{y}=\int-y \cdot \frac{1}{y} \mathrm{~d} y+\mathrm{c}$
$\Rightarrow \frac{x}{y}=-y+\mathrm{c}$ ....(i)
Since $y(1)=1$, i.e., $y=1$ when $x=1$
$\therefore \quad 1=-1+c \Rightarrow c=2$
$\therefore \quad \frac{x}{y}=-y+2 \quad \ldots[$ From (i) $]$
Putting $x=-3$, we get
$-\frac{3}{y}=-y+2$
$\begin{aligned} & \Rightarrow y^2-2 y-3=0 \\ & \Rightarrow(y-3)(y+1)=0\end{aligned}$
Since $y(x)>0, y=3$