Answer
$y=(\dfrac{2}{5x}+cx^{4})^{-1/2}$
Work Step by Step
Here, we have $x^2y'+2xy=y^{3}$
Compare the above equation with the Bernoulli equation with $n=3$
Plug $u=y^{1-(3)}=y^{-2}$ or, $y=u^{-1/2}$
Then $\dfrac{dy}{dx}=(\dfrac{-1}{2})u^{-3/2}\dfrac{du}{dx}$
or, $\dfrac{du}{dx}-\dfrac{4}{x}u=-\dfrac{2}{x^2}$
The integrating factor is: $e^{\int (\frac{-4}{x})dx}=x^{-4}$
In order to determine the general solution, multiply equation (1) with the integrating factor and integrate both sides.
$\int [ux^{-4}]'=(-2) \int x^{-6} dx$
or, $u=\dfrac{2}{5x}+cx^{4} \implies y^{-2}=\dfrac{2}{5x}+cx^{4}$
Thus, $y=(\dfrac{2}{5x}+cx^{4})^{-1/2}$
