The general solution of the differential equation $$(y^2 - x^3)dx - xy\,dy = 0$$, $$(x \neq 0)$$ is (where c is a constant of integration)

We have the differential equation

$$ (y^{2}-x^{3})\,dx-\;x\,y\,dy=0,\qquad (x\neq 0). $$

First we isolate the differentials so that a single derivative appears. Dividing the entire equation by $$dx$$ gives

$$ y^{2}-x^{3}-x\,y\,\frac{dy}{dx}=0. $$

So

$$ x\,y\,\frac{dy}{dx}=y^{2}-x^{3}. $$

Hence

$$ \frac{dy}{dx}=\frac{y^{2}-x^{3}}{x\,y}. $$

To remove the square root-type term in the denominator, we set

$$ z=y^{2}. $$

For this substitution we need the derivative of $$z$$ with respect to $$x$$. Because $$z=y^{2}$$, differentiating both sides gives

$$ \frac{dz}{dx}=2\,y\,\frac{dy}{dx}\;\;\Longrightarrow\;\; y\,\frac{dy}{dx}=\frac12\,\frac{dz}{dx}. $$

Now multiply the earlier equation $$x\,y\,\dfrac{dy}{dx}=y^{2}-x^{3}$$ by $$1$$ and write $$y\,\dfrac{dy}{dx}$$ in terms of $$\dfrac{dz}{dx}$$:

$$ x\left( y\,\frac{dy}{dx}\right)=y^{2}-x^{3}\;\;\Longrightarrow\;\; x\left(\frac12\,\frac{dz}{dx}\right)=z-x^{3}. $$

Simplifying, we obtain

$$ \frac{dz}{dx}= \frac{2z}{x}-2x^{2}. $$

The last expression is a first-order linear ordinary differential equation in $$z$$:

$$ \frac{dz}{dx}-\frac{2}{x}\,z=-2x^{2}. $$

For a linear equation of the form $$\dfrac{dz}{dx}+P(x)\,z=Q(x)$$ the integrating factor is $$\displaystyle \mu(x)=e^{\int P(x)\,dx}.$$ Here $$P(x)=-\dfrac{2}{x}$$, so

$$ \mu(x)=e^{\int -\frac{2}{x}\,dx}=e^{-2\ln x}=x^{-2}. $$

Multiplying the entire differential equation by this integrating factor $$x^{-2}$$ gives

$$ x^{-2}\,\frac{dz}{dx}-2x^{-3}\,z=-2. $$

The left-hand side is the derivative of the product $$z\,x^{-2}$$, because

$$ \frac{d}{dx}\left(z\,x^{-2}\right)=x^{-2}\,\frac{dz}{dx}+z\,\frac{d}{dx}(x^{-2}) =x^{-2}\,\frac{dz}{dx}-2x^{-3}\,z. $$

Thus we can rewrite the equation as

$$ \frac{d}{dx}\bigl(z\,x^{-2}\bigr)=-2. $$

Integrating both sides with respect to $$x$$, we get

$$ z\,x^{-2}=-2\int 1\,dx + C =-2x + C, $$

where $$C$$ is the constant of integration.

Now multiply by $$x^{2}$$ to solve for $$z$$:

$$ z=-2x^{3}+C\,x^{2}. $$

Finally recall that $$z=y^{2}$$, so

$$ y^{2}=-2x^{3}+C\,x^{2}. $$

Gathering all terms on one side and replacing the arbitrary constant $$C$$ by $$c$$ (its sign can be absorbed into the constant), we have

$$ y^{2}+2x^{3}+c\,x^{2}=0. $$

This matches Option D.

Hence, the correct answer is Option D.