Exact Equations & Substitutions · Hard problem

The equation $(6 x y + 4)\,dx + (3 x^{2} + 4 y)\,dy = 0$ is exact. Its solution through $(0, -2)$ is $\psi(x, y) = C$ for a potential $\psi$ with $\psi(0, 0) = 0$. Find $C$.