1. **Problem Statement:** Solve the differential equation $$(x^2 y - 2 x y^2) \, dx - (x^3 - 3 x^2 y) \, dy = 0.$$\n\n2. **Identify the type of equation:** This is a first-order differential equation in the form $$M(x,y) \, dx + N(x,y) \, dy = 0,$$ where $$M = x^2 y - 2 x y^2$$ and $$N = -(x^3 - 3 x^2 y) = -x^3 + 3 x^2 y.$$\n\n3. **Check if the equation is exact:** An equation is exact if $$\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}.$$\nCalculate partial derivatives:\n$$\frac{\partial M}{\partial y} = \frac{\partial}{\partial y}(x^2 y - 2 x y^2) = x^2 - 4 x y,$$\n$$\frac{\partial N}{\partial x} = \frac{\partial}{\partial x}(-x^3 + 3 x^2 y) = -3 x^2 + 6 x y.$$\nSince $$x^2 - 4 x y \neq -3 x^2 + 6 x y,$$ the equation is not exact.\n\n4. **Check for an integrating factor:** Sometimes an integrating factor depending on $x$ or $y$ can make the equation exact.\nCalculate $$\frac{\frac{\partial M}{\partial y} - \frac{\partial N}{\partial x}}{N} = \frac{(x^2 - 4 x y) - (-3 x^2 + 6 x y)}{-x^3 + 3 x^2 y} = \frac{4 x^2 - 10 x y}{-x^3 + 3 x^2 y}.$$\nSimplify numerator and denominator by factoring $x$:\n$$\frac{4 x^2 - 10 x y}{-x^3 + 3 x^2 y} = \frac{2 x (2 x - 5 y)}{x^2 (-x + 3 y)} = \frac{2 (2 x - 5 y)}{x (-x + 3 y)}.$$\nThis expression depends on both $x$ and $y$, so no simple integrating factor depending only on $x$ or $y$ is apparent.\n\n5. **Try substitution or rearrangement:** Notice the structure of $M$ and $N$ suggests homogeneity or a substitution involving $v = \frac{y}{x}$.\nRewrite the equation in terms of $v$:\nSet $$y = v x,$$ then $$dy = v \, dx + x \, dv.$$\nSubstitute into the original equation and simplify to find a separable equation in $v$ and $x$.\n\n6. **Summary of method selection:**\n- Check if the equation is exact by comparing partial derivatives.\n- If not exact, check for integrating factors depending on $x$ or $y$.\n- If integrating factors are not obvious, look for substitutions such as homogeneous substitutions ($v = y/x$) or other methods like separable or linear equations.\n\n**Final answer:** The problem requires substitution $v = \frac{y}{x}$ to solve further, as it is not exact and no simple integrating factor is found.\n\n