\begin{equation}
(3 x^2 y + 2xy + y^3) + (x^2 + y^2) y′= 0\label{A}
\end{equation}
Let $M(x,y) = 3(x^2)y + 2xy + y^3$, $N(x,y) = x^2 + y^2$. Then $M_y(x,y) = 3x^2 + 2x + 3y^2$, $N_x(x,y) = 2x$.
Equation (\ref{A}) is not exact. Lets try to find an integrating factor $\mu=\mu(x)$ to make it exact.
$d\mu/dx = (M_y - N_x)\mu / N\implies
d\mu/dx = (3x^2 + 2x + 3y^2 - 2x)\mu / N \implies
d\mu/dx = 3(x^2 + y^2) \mu / (x^2 + y^2)\implies
d\mu/dx = 3μ\implies
d\mu/ \mu = 3 dx\implies
\ln \mu = 3x \implies
\mu = e^{3x} $
Now multiply the equation(\ref{A}) by $\mu = e^{3x}$
\begin{equation}
\bigl((3 x^2y + 2xy +y^3)e^{3x}\bigr)+ \bigl((x^2+y^2)e^{3x}\bigr)y′= 0
\label{B}
\end{equation}
Now $M(x,y) = 3(x^2y + 2xy +y^3)e^{3x}$, $N(x,y) =(x^2+y^2)e^{3x}$. Then $M_y(x,y) = (3 x^2 + 2x + 3y^2)e^{3x}$, $N_x(x,y) = (2x3+x^2) + 3y^2)e^{3x}$,
$M_y(x,y) = N_x(x,y)$. Therefore the equation is exact. No need to check: it is exact due to construction of $\mu$. V.I.
There is a $\Psi(x, y)$ such that:
\begin{gather}
\Psi _x(x, y) = M(x,y) =3(x^2y + 2xy +y^3)e^{3x},\label{C}\\
\Psi _y (x, y) = N(x,y) = (x^2+y^2)e^{3x}.\label{D}
\end{gather}
Integrating (\ref{C}) we have $\Psi (x, y) = (x^2+ \frac{1}{3} y^3)e^{3x} + f(y)$. Using this, differentiate to get
$\Psi_y (x, y) =(x^2+y^2)e^(3x) + f'(y) $. Easier to start from (\ref{D}) V.I.
Compare this with (\ref{D}): $f'(y) = 0$. $Meaning f(y) = C$, where $C$ is some constant
So,
\begin{equation}
\Psi (x, y) = (x^2+ \frac{1}{3} y^3)e^{3x} +C=0
\end{equation}$ is a solution.
