Exact differential equations;

Find an implicit solution for the differential equation
y(2x+y^3)-x(2x-y^3)dy/dx.

The differential equation is inexact and an integrating factor which is a function of y only is need to solve the problem.

What you show is not an equation at all.

I assume you mean: y(2x + y³) − x(2x − y³) dy/dx = 0
or y(2x + y³) dx − x(2x − y³) dy = 0
which if of the form M(x,y) dx + N(x,y) dy = 0

To find integrating factor, first we find ∂M/∂y and ∂N/∂x
M(x,y) = y(2x+y³) = 2xy + y⁴ -------> ∂M/∂y = 2x + 4y³
N(x,y) = −x(2x−y³) = xy³ − 2x² -----> ∂N/∂x = y³ − 4x

First we check if (∂M/∂y − ∂N/∂x) / N is function of x only
(∂M/∂y − ∂N/∂x) / N
= (2x + 4y³ − y³ + 4x) / (−x(2x−y³))
= (6x + 3y³) / (−x(2x−y³))
= 3 (2x + y³) / (−x(2x − y³)) ---> cannot be simplified.

Next we check that (∂N/∂x − ∂M/∂y) / M is function of y only
(∂N/∂x − ∂M/∂y) / M
= (y³ − 4x − 2x − 4y³) / (y(2x+y³))
= (−6x − 3y³) / (y(2x+y³))
= −3(2x + y³) / (y(2x+y³))
= −3/y

Now we find integrating factor
μ = e^(∫(−3/y) dy) = e^(−3ln(y)) = 1/e^(3ln(y)) = 1/e^(ln(y³)) = 1/y³

Multiply both sides of differential by 1/y³

y(2x + y³) dx − x(2x − y³) dy = 0
1/y² (2x + y³) dx − x/y³ (2x − y³) dy = 0

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Differential equation is now:

(2x/y² + y) dx + (−2x²/y³ + x) dy = 0

Now we check that it is exact:
M(x,y) = 2x/y² + y -----> ∂M/∂y = −4x/y³ + 1
N(x,y) = −2x²/y³ + x -----> ∂N/∂x = −4x/y³ + 1

Differential equation is indeed exact.

Solution is of the form f(x,y) = C
where ∂f/∂x = M(x,y) and ∂f/∂y = N(x,y)

f(x,y) = ∫ M(x,y) dx
f(x,y) = ∫ (2x/y² + y) dx
f(x,y) = x²/y² + xy + θ(y)
where θ(y) is a function of y only, since y is considered constant and ∂/∂x θ(y) = 0

Now we differentiate f(x,y) with respect to y and compare to N(x,y)
∂f/∂y = N(x,y)
−2x²/y³ + x + θ'(y) = −2x²/y³ + x
θ'(y) = 0
θ(y) = 0

Solution: x²/y² + xy = C