If a curve passes through the point (1,-2)

$\because$ Slope of the tangent $=\frac{x^{2}-2 y}{x}$

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

$\frac{d y}{d x}+\frac{2}{x} y=x$

I.F, $=e^{\int^{\frac{1}{4 x}}}=e^{2 \ln x}=x^{2}$

Solution of equation

$y \cdot x^{2}=\int x \cdot x^{2} d x$

$x^{2} y=\frac{x^{4}}{4}+C$

$\therefore$ curve passes through point $(1,-2)$

$(1)^{2}(-2)=\frac{1^{4}}{4}+C$

$\Rightarrow \quad C=\frac{-9}{4}$

Then, equation of curve

$y=\frac{x^{2}}{4}-\frac{9}{4 x^{2}}$

Since, above curve satisfies the point.

Hence, the curve passes through $(\sqrt{3}, 0)$.