Question 1184076
{{{M(x,y) = y - 2y^3}}}  and {{{N(x,y) = -y^5 - 2x}}}


After division by {{{y^3}}},


∂({{{M(x,y)/y^3}}})/∂y ={{{ -2/y^3 }}}= ∂({{{N(x,y)/y^3)}}})/∂x,


and the new differential form becomes exact.