1. The problem is to solve the differential equation: $(1 - xy + x^2 y^2) dx = (x^2 - x^3 y) dy$. 2. Rewrite the equation as $ (1 - xy + x^2 y^2) dx - (x^2 - x^3 y) dy = 0 $. 3. Let $M = 1 - xy + x^2 y^2$ and $N = -(x^2 - x^3 y) = -x^2 + x^3 y$. 4. Check if the differential equation is exact by verifying if $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$. 5. Calculate $\frac{\partial M}{\partial y} = \frac{\partial}{\partial y}(1 - xy + x^2 y^2) = -x + 2x^2 y$. 6. Calculate $\frac{\partial N}{\partial x} = \frac{\partial}{\partial x}(-x^2 + x^3 y) = -2x + 3x^2 y$. 7. Since $\frac{\partial M}{\partial y} \neq \frac{\partial N}{\partial x}$, the equation is not exact. 8. Find an integrating factor to make it exact; try an integrating factor $\mu(y) = y^n$ or $\mu(x) = x^m$. 9. Alternatively, try dividing or multiplying to simplify; see if $\frac{M_y - N_x}{N}$ depends solely on $x$ or $y$ for an integrating factor. 10. Calculate $\frac{M_y - N_x}{N} = \frac{(-x + 2x^2 y) - (-2x + 3x^2 y)}{-x^2 + x^3 y} = \frac{(-x + 2x^2 y + 2x - 3x^2 y)}{-x^2 + x^3 y} = \frac{(x - x^2 y)}{-x^2 + x^3 y} = \frac{x(1 - x y)}{-x^2(1 - x y)} = \frac{x}{-x^2} = -\frac{1}{x}$. 11. Since this depends only on $x$, the integrating factor is $\mu(x) = e^{\int -\frac{1}{x} dx} = e^{-\ln|x|} = x^{-1}$. 12. Multiply both $M$ and $N$ by $x^{-1}$: $\tilde M = x^{-1} M = x^{-1} - y + x y^2$, $\tilde N = x^{-1} N = -x + x^2 y$. 13. Check exactness again: $\frac{\partial \tilde M}{\partial y} = -1 + 2 x y$, and $\frac{\partial \tilde N}{\partial x} = -1 + 2 x y$. 14. This shows the equation is now exact. 15. Integrate $\tilde M$ with respect to $x$: $\int (x^{-1} - y + x y^2) dx = \int x^{-1} dx - \int y dx + \int x y^2 dx = \ln|x| - x y + \frac{1}{2} x^2 y^2 + h(y)$. 16. Differentiate this w.r.t. $y$: $\frac{\partial}{\partial y} [\ln|x| - x y + \frac{1}{2} x^2 y^2 + h(y)] = -x + x^2 y + h'(y)$. 17. This must equal $\tilde N = -x + x^2 y$. 18. Hence, $h'(y) = 0$, so $h(y)$ is constant. 19. Therefore, the implicit solution is: $$ \ln|x| - x y + \frac{1}{2} x^2 y^2 = C $$. 20. This is the general solution to the differential equation.