Answer
$y=(1+\dfrac{c}{x^3})^{1/3}$
Work Step by Step
Here, we have $xy'+y=y^{-2}$
Compare the above equation with the Bernoulli equation with $n=-2$
Plug $u=y^{1-(-2)}=y^{3}$ or, $y=u^{1/3}$
Then $\dfrac{dy}{dx}=(\dfrac{1}{3})u^{-2/3}\dfrac{du}{dx}$
or, $\dfrac{du}{dx}+\dfrac{3}{x}u=\dfrac{3}{x}$
The integrating factor is: $e^{\int (\frac{3}{x})dx}=x^{3}$
In order to determine the general solution, multiply equation (1) with the integrating factor and integrate both sides.
$\int [x^3u]'=\int 3x^2 dx$
or, $u=1+cx^{-3} \implies y^3=1+cx^{-3}$
Thus, $y=(1+\dfrac{c}{x^3})^{1/3}$