We have,
$$
x^2 y d x-\left(x^3+y^3\right) d y=0 \Rightarrow \frac{d y}{d x}=\frac{x^2 y}{x^3+y^3}
$$

Given differentiate equation is in the form of homogeneous differentiate equation.
So, let $y=v x$
$$
\begin{aligned}
& \Rightarrow \quad \frac{d y}{d x}=v+x \frac{d v}{d x} \\
& \therefore \quad v+x \frac{d v}{d x}=\frac{v}{1+v^3} \\
& \Rightarrow x \frac{d v}{d x}=\frac{v}{1+v^3}-v \quad \Rightarrow x \frac{d v}{d x}=\frac{v-v-v^4}{1+v^3} \\
& \Rightarrow x \frac{d v}{d x}=\frac{-v^4}{1+v^3} \quad \Rightarrow \frac{1+v^3}{v^4} d v=-\frac{d x}{x}
\end{aligned}
$$


Integrating both side, we get
$$
\begin{aligned}
& \int\left(\frac{1}{v^4}+\frac{1}{v}\right) d v=-\int \frac{1}{x} d x \Rightarrow-\frac{1}{3 v^3}+\log |v|=-\log x+c \\
& \Rightarrow-\frac{x^3}{3 y^3}+\log \left|\frac{y}{x}\right|=-\log |x|+c \Rightarrow-\frac{x^3}{3 y^3}+\log |y|=c
\end{aligned}
$$