Solve \begin{align*}
(2xy^2 + 2y) + (2x^2y + 2x)y' &= 0\\
\implies (2xy^2 + 2y)dx + (2x^2y + 2x)dy &= 0
\end{align*}
Let $M = 2xy^2 + 2y$ and $N = 2x^2y + 2x$
\begin{equation}
M_y = \frac{d M}{dy} = 4xy + 2x\quad\quad\quad N_x = \frac{d N}{dy} = 4xy + 2x
\end{equation}
Since $M_y = N_x$, the equation is exact so $\exists \psi(x, y)\quad s.t \quad \frac{\partial \psi}{\partial x} = M$ and $\frac{\partial \psi}{\partial y} = N$
Therefore
\begin{align*}
\psi(x, y) &= \int M dx \\
&= \int 2x^2y + 2y dx\\
&= x^2y^2 + 2xy + h(y)
\end{align*}
Since
\begin{align*}
\frac{\partial \psi}{\psi y} &= 2x^2y + 2x +h'(y)\\
\implies h'(y) &= 0\\
\implies h(y) &= C
\end{align*}
Therefore, the solution is $x^2y^2 + 2xy = C$
