$\frac{dy}{y^2-2}=xdx$
$\int \frac{dy}{y^2-2}=-\ln(2\sqrt{ 2}y+4)+\ln(2\sqrt{ 2 }y-4)=\ln\left( \frac{2\sqrt{ 2 }y-4}{2\sqrt{ 2 }y+4} \right)=\ln\left( \frac{y-\sqrt{ 2 }}{y+\sqrt{ 2 }} \right)=\ln\left( 1-\frac{2\sqrt{ 2 }}{y+\sqrt{ 2 }} \right)$
$\int xdx = \frac{1}{2}x^2$
$y=\frac{1}{1-e^{1/2x^2}}$
$\frac{dy}{dx}=-\frac{1}{(1-e^{1/2x^2})^2}(-xe^{1/2x^2})$

$y^2-2=(y+\sqrt{ 2 })(y-\sqrt{2})$ 
$\frac{1}{(y+\sqrt{ 2 })(y-\sqrt{ 2 })}=\frac{a}{y+\sqrt{ 2 }}+\frac{b}{y-\sqrt{ 2 }}=-\frac{1}{2\sqrt{ 2 }y+4}+\frac{1}{2\sqrt{ 2 }y-4}$ 
$a(y-\sqrt{ 2 })+b(y+\sqrt{ 2 })=1$
$b=\frac{1}{2\sqrt{ 2 }},a=-\frac{1}{2\sqrt{ 2 }}$