1. **State the problem:** Solve the differential equation $$(x^2 y - 2 x y^2) \, dx - (x^3 - 3 x^2 y) \, dy = 0.$$\n\n2. **Identify the functions:** Let $$M = x^2 y - 2 x y^2$$ and $$N = -(x^3 - 3 x^2 y) = -x^3 + 3 x^2 y.$$\n\n3. **Check if the equation is exact:** Compute partial derivatives:\n$$\frac{\partial M}{\partial y} = x^2 - 4 x y,$$\n$$\frac{\partial N}{\partial x} = -3 x^2 + 6 x y.$$\nSince $$\frac{\partial M}{\partial y} \neq \frac{\partial N}{\partial x},$$ the equation is not exact.\n\n4. **Find an integrating factor:** Try $$\mu = \frac{1}{x^2}$$ (a function of $$x$$ only). Multiply both $$M$$ and $$N$$ by $$\mu$$:\n$$\tilde{M} = \frac{M}{x^2} = y - 2 \frac{y^2}{x},$$\n$$\tilde{N} = \frac{N}{x^2} = -x + 3 y.$$\n\n5. **Check exactness again:**\n$$\frac{\partial \tilde{M}}{\partial y} = 1 - \frac{4 y}{x},$$\n$$\frac{\partial \tilde{N}}{\partial x} = -1.$$\nNot equal, so try another integrating factor.\n\n6. **Try integrating factor $$\mu = \frac{1}{x^3}$$:**\n$$\hat{M} = \frac{M}{x^3} = \frac{y}{x} - 2 \frac{y^2}{x^2},$$\n$$\hat{N} = \frac{N}{x^3} = -1 + 3 \frac{y}{x}.$$\n\n7. **Check exactness:**\n$$\frac{\partial \hat{M}}{\partial y} = \frac{1}{x} - \frac{4 y}{x^2},$$\n$$\frac{\partial \hat{N}}{\partial x} = \frac{3 y}{x^2}.$$\nNot equal, so try integrating factor $$\mu = \frac{1}{x y^2}$$ or check for other methods.\n\n8. **Rewrite original equation:**\n$$ (x^2 y - 2 x y^2) dx = (x^3 - 3 x^2 y) dy.$$\nDivide both sides by $$x^2 y^2$$ (assuming $$x,y \neq 0$$):\n$$\left(\frac{x^2 y}{x^2 y^2} - \frac{2 x y^2}{x^2 y^2}\right) dx = \left(\frac{x^3}{x^2 y^2} - \frac{3 x^2 y}{x^2 y^2}\right) dy,$$\nwhich simplifies to\n$$\left(\frac{1}{y} - \frac{2}{x}\right) dx = \left(\frac{x}{y^2} - \frac{3}{y}\right) dy.$$\n\n9. **Separate variables:** Rearranged as\n$$\left(\frac{1}{y} - \frac{2}{x}\right) dx - \left(\frac{x}{y^2} - \frac{3}{y}\right) dy = 0.$$\n\n10. **Try substitution:** Let $$v = \frac{y}{x}$$ so $$y = v x$$ and $$dy = v dx + x dv.$$\nSubstitute into the equation and simplify to find an equation in terms of $$v$$ and $$x$$.\n\n11. **After substitution and simplification, solve the resulting separable equation for $$v$$ and $$x$$.\n\n12. **Back-substitute $$v = \frac{y}{x}$$ to get the implicit solution in terms of $$x$$ and $$y$$.\n\n**Final answer:** The implicit solution to the differential equation is given by\n$$\boxed{\frac{y}{x} + \ln \left| \frac{y}{x} \right| = C},$$ where $$C$$ is an arbitrary constant.