$xy \, y' = 1 + y^2$
$xy \dfrac{dy}{dx} = 1 + y^2$
$\dfrac{y \, dy}{1 + y^2} = \dfrac{dx}{x}$
$\displaystyle \frac{1}{2} \int \dfrac{2y \, dy}{1 + y^2} = \int \dfrac{dx}{x}$
$\frac{1}{2} \ln (1 + y^2) = \ln x + \frac{1}{2} \ln c$
$\ln (1 + y^2) = 2\ln x + \ln c$
$\ln (1 + y^2) = \ln x^2 + \ln c$
$\ln (1 + y^2) = \ln cx^2$
$1 + y^2 = cx^2$
when $x = 2$, $y = 3$
$1 + 3^2 = c(2^2)$
$c = \frac{10}{4} = \frac{5}{2}$
then,
$1 + y^2 = \frac{5}{2}x^2$
$y^2 = \frac{5}{2}x^2 - 1$
$y^2 = \dfrac{5x^2 - 2}{2}$
$y = \sqrt{\dfrac{5x^2 - 2}{2}}$
$y = \sqrt{\dfrac{5x^2 - 2}{2} \cdot \dfrac{2}{2}\,}$
$y = \sqrt{\dfrac{10x^2 - 4}{4}}$
$y = \frac{1}{2} \sqrt{10x^2 - 4}$ answer
