Question 1196937
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<pre>

{{{(xdy-ydx)/y^2}}} = {{{x^3dx}}}


Left side is  {{{d(-(x/y))}}}.


Right side is  {{{d((1/4)*x^4)}}}.


So, your equation is

    {{{d(-(x/y))}}} = {{{d((1/4)*x^4)}}}.


It implies

    {{{-(x/y)}}} = {{{(1/4)x^4}}} + C,    where  C = const.


It means that 

    4x + x^4*y = Cy.


which is your answer.
</pre>

Solved.