Determine whether the given DE is exact. If it is exact, solve it.
$({x^3} + {y^3})dx + 3x{y^2}dy = 0$

Hint: First identify M and N from the given equation, then differentiate M with respect to y and differentiate N with respect to x, then observe that they are equal or not. If they are equal then solve the equation putting integration on M for x here take y as constant and integrate N for y, take x as constant or say not taking the terms involving x.

Formula Used:
The differential equation $Mdx + Ndy = 0$ is exact iff
$\dfrac{{\partial M}}{{\partial y}} = \dfrac{{\partial N}}{{\partial x}}$ ,
then the solution is $\int\limits_{y = constant} {Mdx} + \int\limits_{terms{\rm{ without x}}} {Ndy} = 0$

Complete step by step solution:
Here, $M = {x^3} + {y^3}$ and $N = 3x{y^2}$
So, $\dfrac{{\partial M}}{{\partial y}} = 3{y^2}$and
$\dfrac{{\partial N}}{{\partial x}} = 3{y^2}$
Therefore, $\dfrac{{\partial M}}{{\partial y}} = \dfrac{{\partial N}}{{\partial x}}$.
Hence, the equation is exact.
Now, the solution is,
$\int\limits_{y = constant} {Mdx} + \int\limits_{terms{\rm{ without x}}} {Ndy} = 0$
$\int\limits_{y = constant} {({x^3} + {y^3})dx} + \int\limits_{terms{\rm{ without x}}} {0.dy} = 0$
$\dfrac{{{x^4}}}{4} + x{y^3} = C$ , where C is an integrating constant.

Hence the required answer is $\dfrac{{{x^4}}}{4} + x{y^3} = C$

Note: Students sometimes get confused and differentiate M with respect to x and differentiate N with respect to y to check the exactness and leads to an incorrect solution. The condition to check the exactness of a differential equation is $\dfrac{{\partial M}}{{\partial y}} = \dfrac{{\partial N}}{{\partial x}}$.