Question

(x^3+y^3)=(3xy^2)dy/dx

Answer 1

A Bernoulli differential equation

$$\dfrac{dy}{dx}-\dfrac{1}{3x}y=\dfrac{x^2}{3}y^{-2}$$

Substitute

$$z=y^{1-(-2)}=y^3$$

$$\dfrac{dz}{dx}=3y^2\dfrac{dy}{dx}$$

$$\dfrac{dy}{dx}=\dfrac{1}{3y^2}\dfrac{dz}{dx}$$

Then

$$\dfrac{dz}{dx}-\dfrac{1}{x}z=x^2$$

$$IF=e^{\int (-1/x)dx}=\dfrac{1}{x}$$

$$\dfrac{1}{x}\dfrac{dz}{dx}-\dfrac{1}{x^2}z=x$$

$$\dfrac{d}{dx}(\dfrac{z}{x})=x$$

$$\dfrac{z}{x}=\dfrac{x^2}{2}+C$$

$$z=\dfrac{x^3}{2}+Cx$$

$$y^3=\dfrac{x^3}{2}+Cx$$