Question 1184041
Next time, try to make proper use of parentheses. I will assume that your D.E. is

{{{dy/dx = (-x^4 -2y)/(2x + 4y^2)}}}.


<===> {{{(x^4 +2y)dx + (2x + 4y^2)dy=0}}}, with {{{M(x,y) = x^4 +2y}}} and {{{N(x,y) =2x + 4y^2}}}.

As {{{M[y] = 2 = N[x]}}}, the D.E. is exact. 

As such, you can retrieve the potential function F(x,y) by setting ∂F/∂x = M(x,y) and ∂F/∂y = N(x,y).


In the end what you will get is {{{F(x,y) = x^5/5 + 2yx + (4/3)y^3 = C }}}

Can you see now where you went wrong?