Given differential equation can be rewritten as
$$
\frac{d y}{d x}=\frac{x^3+y^3}{x y^2}
$$
It is a homogeneous differential equation.
Put
$$
y=v x \Rightarrow \frac{d y}{d x}=v+x \frac{d v}{d x}
$$
$$
\begin{array}{rlrl}
& \therefore & x \frac{d v}{d x}+v & =\frac{x^3+v^3 x^3}{x^3 v^2} \\
\Rightarrow & x \frac{d v}{d x}+v & =\frac{1+v^3}{v^2} \\
\Rightarrow & x \frac{d v}{d x} & =\frac{1}{v^2} \\
\Rightarrow & v^2 d v & =\frac{d x}{x}
\end{array}
$$
On integrating both sides, we get
$$
\begin{aligned}
\frac{v^3}{3} & =\log x+\log c \\
\Rightarrow \quad \frac{1}{3}\left(\frac{y}{x}\right)^3 & =\log x+\log c \\
\Rightarrow \quad y^3 & =3 x^3 \log c x
\end{aligned}
$$