= Solution
The original <differential equation> is defined only where $x\ne0$ and $y\ne0$. Put $w=y^3$, so the <chain rule> gives $w'=3y^2y'$. Multiplying the equation by $3y^2$ yields a first-order <linear ordinary differential equation>:
$$
w'=2x^2+\frac wx,\qquad w'-\frac wx=2x^2.
$$
An <integrating factor> on either half-line is $1/x$, up to an irrelevant nonzero constant. Equivalently, apply the <quotient rule> directly:
$$
\left(\frac wx\right)'=\frac{w'-w/x}{x}=2x.
$$
Integrating gives $w/x=x^2+C$. Thus the real <general solution> is
$$
\boxed{y(x)=\sqrt[3]{x^3+Cx},\qquad C\in\mathbb R,}
$$
on any connected interval where $x\ne0$ and $x^3+Cx\ne0$. These restrictions preserve both the original denominator and differentiability of the real cube root. In particular $C=0$ includes $y=x$ on either half-line; this solution has not been lost through division by an expression involving $y/x-1$.
For verification, differentiating $y^3=x^3+Cx$ gives $3y^2y'=3x^2+C$, while $(2x^3+y^3)/x=3x^2+C$. The two sides agree wherever the original equation is defined.
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