Answer
An integrating factor is $\mu(x)= e^{3x} $
So
The differential equation :
$$
\left(3 x^{2} y+2 x y+y^{3}\right) d x+\left(x^{2}+y^{2}\right) d y=0
$$
becomes exact when multiplied by the integrating factor $\mu(x)= e^{3x} $ and its solutions are given implicitly by
$$
e^{3 x} (3yx^{2}+y^{3}) =c
$$
where $ c $ is an arbitrary constant
Work Step by Step
$$
\left(3 x^{2} y+2 x y+y^{3}\right) d x+\left(x^{2}+y^{2}\right) d y=0 \quad \quad (i)
$$
Comparing this Equation with the differential form:
$$
M(x,y) d x+N(x,y) d y=0
$$
we observe that
$$
\begin{aligned}M(x,y) &=\left(3 x^{2} y+2 x y+y^{3}\right)
\\ N(x,y) &=\left(x^{2}+y^{2}\right) \end{aligned}
$$
By calculating $M_{y}$ and $N_{x}$ , we find that
$$
\begin{aligned} M_{y}(x, y) &= 3 x^{2}+2 x +3 y^{2}
\\ N_{x}(x, y) & = 2x \end{aligned}
$$
we find that
$$
\frac{M_{y}(x, y)-N_{x}(x, y)}{N(x, y)}=\frac{\left(3 x^{2}+2 x +3 y^{2} \right) -2x }{\left(x^{2}+y^{2}\right) }=3.
$$
Thus there is an integrating factor $μ $ that is a function of $x$ only, and it satisfies the differential equation
$$
\frac{d \mu}{d x}=3 \mu .
$$
Hence
$$
\frac{d \mu}{ \mu}=3 dx
$$
Integrating, we obtain
$$
\mu=e^{3x}
$$
Multiplying Eq. (i) by this integrating factor, we obtain
$$
\left(3 x^{2} y+2 x y+y^{3}\right) e^{3x} d x+\left(x^{2}+y^{2}\right)e^{3x} d y=0
$$
The latter equation is exact, since
$$
M_{y}(x, y)=\left(3 x^{2} +2 x+3y^{2 }\right) e^{3x} =N_{x}(x, y).
$$
Thus there is a $\psi (x, y)$ such that
$$
\begin{aligned} \psi_{x}(x, y) &=M(x,y) =\left(3 x^{2} y+2 x y+y^{3} \right) e^{3x} ,
\\
\psi_{y}(x, y) &=N(x,y) =\left(x^{2}+y^{2} \right) e^{3x}
\end{aligned} \quad \quad (ii)
$$
Integrating the first of these equations with respect to $x$ , we obtain
$$
\begin{split}
\psi(x, y) & =\int { \left(3 x^{2} y+2 x y+y^{3} \right) e^{3x} } dx \\
& = 3y \int { x^{2} e^{3x}dx}+2y \int {x e^{3x}dx}+y^{3} \int { e^{3x}dx } +h(y)
\\ & = 3y (\frac{1}{3} e^{3 x} x^{2}-\frac{2}{27}\left(3 e^{3 x} x-e^{3 x}\right))
+2y (\frac{1}{9}\left(e^{3 x} \cdot 3 x-e^{3 x}\right)) +y^{3} (\frac{1}{3} e^{3 x}) +h(y)
\\ & = e^{3 x} (yx^{2}+\frac{1}{3}y^{3})+h(y)
\quad \quad \quad\quad\quad (iii)
\end{split}
$$
where $h(y)$ is an arbitrary function of $y$ only. To try to satisfy the second of Eqs. (ii), we compute $\psi_{y}(x, y) $ from Eq. (iii) and set it equal to $N$, obtaining
$$
\psi_{y}(x, y)= e^{3 x} (x^{2}+y^{2}) +h^{\prime}(y)= \left(x^{2}+y^{2} \right) e^{3x}
$$
Thus $h^{\prime}(y)=0 $ and $h(y)=0$ , the constant of integration can be omitted since any solution of the preceding differential equation is satisfactory.
Now substituting for $h(y)$ in Eq. (iii) gives
$$
\psi(x, y)= e^{3 x} (yx^{2}+\frac{1}{3}y^{3})
$$
Hence solutions of Eq. (i) are given implicitly by
$$
\psi(x, y) = e^{3 x} (yx^{2}+\frac{1}{3}y^{3}) =c_{1}
$$
where $ c_{1} $ is an arbitrary constant.
The solution can be rewritten as follows
$$
\psi(x, y) = e^{3 x} (3yx^{2}+y^{3}) =c
$$
where $ c=3c_{1} $ is an arbitrary constant
