$x y \frac{d y}{d x}=2 y^2-x^2$
$\frac{d y}{d x}=\frac{2 y}{x}-\frac{x}{y}$
$\frac{d y}{d x}-\frac{2 y}{x}=\frac{-x}{y} \Rightarrow y \frac{d y}{d x} \frac{-2 y^2}{x}=-x$ $\ldots$ (i)
Put, $\quad v=y^2$ $\ldots$ (ii)
$\Rightarrow\left(\frac{d v}{d x}=2 y \frac{d y}{d x}\right) \Rightarrow\left(\frac{1}{2} \frac{d v}{d x}=y \frac{d y}{d x}\right)$
From Eq. (i) $\frac{1}{2} \frac{d v}{d x}-\frac{2 v}{x}=-x$
$\frac{d v}{d x}-\frac{4 v}{x}=-2 x$
$\mathrm{IF}=e^{\int p d x}=e^{\int-\frac{4}{x} d x}=e^{-4 \log x}$
$=e^{\log x^{-4}}=x^{-4}$
Complete solution is
$x^{-4} \cdot v=\int(-2 x) \cdot x^{-4} d x+c$
$\frac{v}{x^4}=-2 \int \frac{d x}{x^3}+c \Rightarrow \frac{v}{x^4}=\frac{1}{x^2}+c$
$v=x^2+c x^4$
$y^2=c x^4+x^2 \quad[$ from Eq. (ii) $]$
which is the required family of curves.