Answer
$y=(1+\dfrac{c}{x^3})^{1/3}$
Work Step by Step
On comparing the equation $xy'+y=y^{-2}$ with Bernoulli with $n=-2$
Consider $p=y^{1-(-2)}=y^{3}$
Then $\dfrac{dy}{dx}=(\dfrac{p^{-2/3}}{3})\dfrac{dp}{dx}$
$\implies \dfrac{dp}{dx}+\dfrac{3p}{x}=\dfrac{3}{x}$
The integrating factor is: $I=e^{\int (\frac{3}{x})dx}=x^{3}$
$x^{3}[\dfrac{dp}{dx}+\dfrac{3p}{x}]=x^{3}(\dfrac{3}{x})$
This gives: $\int [x^3u]'=\int 3x^2 dx$
$\implies p=1+cx^{-3}$
Thus, $y^3=1+cx^{-3} \implies y=(1+\dfrac{c}{x^3})^{(1/3)}$
