Answer
$\dfrac{1}{3} \ln |y^3-2|= x^3+c$
Work Step by Step
Given: $y^2 \dfrac{dy}{dx}=3x^2y^3-6x^2$
Re-arrange the given equation and integrate as follows:.
Then $ \int 3x^2 dx =\int \dfrac{y^2}{(y^3-2)}dy\implies \int 3x^2 dx =(\dfrac{1}{3}) \int \dfrac{3y^2}{(y^3-2)}dy$ ....(1)
As we know that $\int x^n dx=\dfrac{x^{n+1}}{n+1}+c$
Thus, $x^3+c =(\dfrac{1}{3}) \int \dfrac{3y^2}{(y^3-2)}dy$ ...(2)
Let us take the help of $u$ substitution.
plug $u=y^3 \implies du =3y^2 dy$
Equation (2) becomes: $\dfrac{1}{3} \int \dfrac{du}{u-2}=x^3+c \implies \dfrac{1}{3} \ln |u-2|= x^3+c$
Hence, $\dfrac{1}{3} \ln |y^3-2|= x^3+c$
