Q:

Can you solve this Lagrange differential equation?

How can i found the solution for this: 
solve this using Lagrange method (differential equation): 
$$\color{red}{y=2y'x+y^3}$$
Hint: $$\color{#00F}{y^3 \equiv y'''}$$
Can you explain the method?
This is a capture for this example with solution. 

$$\color{#00F}{Translation:}$$
Example:
Find the general solution for this differential equation
$$\color{#F00}{y=2px+p^3}$$
Solution:
This equation is a differential Lagrange equation, by integrate the 02 (two) parts we obtain:  
$$\color{#080}{p=\frac{dy}{dx}=2p+2x\frac{dp}{dx}+3p^2\frac{dp}{dx}}$$

A:

We are given (I think there is an issue in the hint):
$$\tag 1 y = 2y' x + y^3, y^3 \equiv (y')^3$$
If we let $p = \dfrac{dy}{dx}$, we can rewrite $(1)$ as:
$$\tag 2 y = 2px + p^3$$
Differentiating $(2)$ wrt $x$ yields:
$$\dfrac{dy}{dx} = p = 2p + 2 x \dfrac{dp}{dx} + 3 p^2 \dfrac{dp}{dx}$$
This reduces to:
$$\tag 3 3p^2\frac{dp}{dx} + 2x\frac{dp}{dx} + p = 0$$
Solving $(3)$ yields four solutions as:
$$p(x) = \pm ~\sqrt{\frac{2}{3}} \sqrt{\pm ~\sqrt{c_1+x^2}-x}$$
From our initial substitution, we have $p = \dfrac{dy}{dx}$, so we can write:
$$\dfrac{dy}{dx} = \pm ~\sqrt{\frac{2}{3}} \sqrt{\pm ~\sqrt{c_1+x^2}-x}$$
We can solve these four (it is possible that not all four satisfy the ODE) for $y(x)$ by rearranging and integrating. For example:
$$\int dy = -\sqrt{\frac{2}{3}} \int \sqrt{~-\sqrt{c_1+x^2}-x}~~dx $$
where $c_1$ is a constant.

