1. **State the problem:** Solve the differential equation $$(3x^2 y - x^2) \, dx + dy = 0.$$ 2. **Rewrite the equation:** Express it in the form $$M(x,y) \, dx + N(x,y) \, dy = 0,$$ where $$M = 3x^2 y - x^2$$ and $$N = 1.$$ 3. **Check if the equation is exact:** Calculate $$\frac{\partial M}{\partial y} = 3x^2$$ and $$\frac{\partial N}{\partial x} = 0.$$ Since $$\frac{\partial M}{\partial y} \neq \frac{\partial N}{\partial x},$$ the equation is not exact. 4. **Find an integrating factor:** Since $$\frac{\partial M}{\partial y} - \frac{\partial N}{\partial x} = 3x^2,$$ which depends only on $$x,$$ try an integrating factor $$\mu(x)$$. 5. **Calculate integrating factor:** $$\mu(x) = e^{\int \frac{\frac{\partial M}{\partial y} - \frac{\partial N}{\partial x}}{N} dx} = e^{\int 3x^2 dx} = e^{x^3}.$$ 6. **Multiply the entire equation by $$\mu(x)$$:** $$e^{x^3}(3x^2 y - x^2) dx + e^{x^3} dy = 0.$$ 7. **Check exactness again:** Let $$\tilde{M} = e^{x^3}(3x^2 y - x^2)$$ and $$\tilde{N} = e^{x^3}.$$ Calculate $$\frac{\partial \tilde{M}}{\partial y} = 3x^2 e^{x^3}$$ and $$\frac{\partial \tilde{N}}{\partial x} = 3x^2 e^{x^3}.$$ They are equal, so the equation is exact now. 8. **Find potential function $$\Psi(x,y)$$:** Integrate $$\tilde{M}$$ with respect to $$x$$: $$\Psi(x,y) = \int e^{x^3}(3x^2 y - x^2) dx + h(y).$$ Split the integral: $$\int e^{x^3} 3x^2 y dx - \int e^{x^3} x^2 dx + h(y).$$ Use substitution $$u = x^3, du = 3x^2 dx$$ for the first integral: $$\int y e^u du = y e^{x^3}.$$ For the second integral, use substitution $$u = x^3, du = 3x^2 dx$$ so $$x^2 dx = \frac{du}{3}$$: $$\int e^u \frac{du}{3} = \frac{1}{3} e^{x^3}.$$ So, $$\Psi(x,y) = y e^{x^3} - \frac{1}{3} e^{x^3} + h(y).$$ 9. **Differentiate $$\Psi$$ with respect to $$y$$:** $$\frac{\partial \Psi}{\partial y} = e^{x^3} + h'(y).$$ Set equal to $$\tilde{N} = e^{x^3}$$: $$e^{x^3} + h'(y) = e^{x^3} \implies h'(y) = 0.$$ 10. **Conclude:** $$h(y) = C,$$ a constant. 11. **General solution:** $$\Psi(x,y) = y e^{x^3} - \frac{1}{3} e^{x^3} = C.$$ **Final answer:** $$y e^{x^3} - \frac{1}{3} e^{x^3} = C.$$