1. The problem gives the differential equation $$x(y^2 + 13) \, dx + y(x^2 + 6) \, dy = 0$$ and asks to analyze it. 2. We are given a potential function candidate: $$\frac{x^2 y^2}{2} + \frac{13 x^2}{2} + \frac{6 y^2}{2} = C$$ where $C$ is a constant. 3. To verify this is a potential function, recall that for an exact differential equation $M(x,y) \, dx + N(x,y) \, dy = 0$, there exists a function $\Psi(x,y)$ such that: $$\frac{\partial \Psi}{\partial x} = M(x,y)$$ $$\frac{\partial \Psi}{\partial y} = N(x,y)$$ 4. Here, $M = x(y^2 + 13)$ and $N = y(x^2 + 6)$. 5. Compute partial derivatives of the potential function candidate: $$\frac{\partial}{\partial x} \left( \frac{x^2 y^2}{2} + \frac{13 x^2}{2} + \frac{6 y^2}{2} \right) = x y^2 + 13 x = M$$ $$\frac{\partial}{\partial y} \left( \frac{x^2 y^2}{2} + \frac{13 x^2}{2} + \frac{6 y^2}{2} \right) = x^2 y + 6 y = N$$ 6. Since these match $M$ and $N$, the given function is indeed a potential function. 7. The implicit solution to the differential equation is: $$\frac{x^2 y^2}{2} + \frac{13 x^2}{2} + \frac{6 y^2}{2} = C$$ This represents the family of curves satisfying the differential equation. Final answer: $$\frac{x^2 y^2}{2} + \frac{13 x^2}{2} + \frac{6 y^2}{2} = C$$