Answer
An integrating factor is $$ \mu (y)=y^{2} $$
the given differential equation:
$$
\left[4\left(x^{3} / y^{2}\right)+(3 / y)\right] d x+\left[3\left(x / y^{2}\right)+4 y\right] d y=0
$$
is not exact but becomes exact when multiplied by the integrating factor $ \mu (y)=y^{2} $ and the solutions are given implicitly by
$$
x^{4}+3 x y +y^{4} =c
$$
where $c$ is an arbitrary constant
Work Step by Step
$$
\left[4\left(x^{3} / y^{2}\right)+(3 / y)\right] d x+\left[3\left(x / y^{2}\right)+4 y\right] d y=0
\quad \quad (i)
$$
Comparing this Equation with the differential form:
$$
M(x,y) d x+N(x,y) d y=0
$$
we observe that
$$
\begin{aligned}M(x,y) &= \left[4\left(x^{3} / y^{2}\right)+(3 / y)\right]
\\ N(x,y) &=\left[3\left(x / y^{2}\right)+4 y\right] \end{aligned}
$$
By calculating $M_{y}$ and $N_{x}$ , we find that
$$
\begin{aligned} M_{y}(x, y) &= \left[-8\left(x^{3} / y ^{3}\right)+(-3 / y^{2})\right]
\\ N_{x}(x, y) & =3/ y^{2} \end{aligned}
$$
we obtain that
$$
M_{y}(x, y) \neq N_{x}(x, y)
$$
so the given equation is not exact
On computing the quantity $ \frac{N_{x}(x, y)-M_{y}(x, y)}{M(x, y)}$ we find
that
$$
\begin{split}
\frac{N_{x}(x, y)-M_{y}(x, y)}{M(x, y)} & =\frac{ 3/ y^{2} -\left[-8\left(x^{3} / y ^{3}\right)+(-3 / y^{2})\right]}{\left[4\left(x^{3} / y^{2}\right)+(3 / y)\right] } \\
& = \frac{ 6/ y^{2} + 8\left(x^{3} / y ^{3}\right)}{\left[4\left(x^{3} / y^{2}\right)+(3 / y)\right] } \\
&=\frac{2}{y}
\end{split}
$$
Thus there is an integrating factor $\mu $ that is a function of $y$ only, and it satisfies the differential equation
$$
\frac{d \mu}{d y}=\mu [\frac{2}{y}]
$$
Hence
$$
\mu (y)=y^{2}
$$
Multiplying Eq. (i) by this integrating factor, we obtain
$$
\left( 4 x^{3}+3 y \right) d x+\left( 3 x +4 y^{3} \right) d y=0
$$
The latter equation is exact, since
$$
M_{y}(x, y)=3=N_{x}(x, y)
$$
Thus there is a $\psi (x, y)$ such that
$$
\begin{aligned} \psi_{x}(x, y) &=M(x,y) =\left( 4 x^{3}+3 y \right) ,
\\ \psi_{y}(x, y) &=N(x,y) =\left( 3 x +4 y^{3} \right) \end{aligned} \quad \quad (ii)
$$
Integrating the first of these equations with respect to $x$ , we obtain
$$
\psi(x, y)=\left( x^{4}+3 x y \right) +h(y) \quad \quad (iii)
$$
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
$$
\psi_{y}(x, y)=-3 x+h^{\prime}(y)=\left( 3 x +4 y^{3} \right)
$$
Thus $h^{\prime}(y)=4y^{3} $ and $h(y)=y^{4}$ 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
$$
\psi(x, y)=\left( x^{4}+3 x y \right) +y^{4}
$$
Hence solutions of Eq. (i) are given implicitly by
$$
\psi(x, y)= x^{4}+3 x y +y^{4} =c
$$
where $c$ is an arbitrary constant
