Exact Equations & Substitutions · Hard problem

The equation $(4 x y + 1)\,dx + (2 x^{2} + 4 y)\,dy = 0$ is exact. Its solution through $(-2, -1)$ is $\psi(x, y) = C$ for a potential $\psi$ with $\psi(0, 0) = 0$. Find $C$.