\begin{equation}
(2xy^2+2y)+(2x^2y+2x)y'=0
\end{equation} write y' into dy/dx, then we get
\begin{equation}
(2xy^2+2y)dx+(2x^2y+2x)dy=0
\end{equation} Let M = 2xy^2+2y , N = 2x^2y+2x , we get
\begin{equation}
\frac{dM}{dy}= 4xy + 2 , \frac{dN}{dx} = 4xy + 2
\end{equation} Since dM/dy = dN/dx, therefore they are exact. So ∃ 𝛗(x,y) = ∫ M dx
\begin{equation}
𝛗(x,y) = ∫ M dx = ∫ 2xy^2 + 2y dx = x^2y^2 + 2xy + h(y)
\end{equation}d𝛗(x,y)/dy = N
\begin{equation}
4x^2y + 2x +h'(y) = 2x^2y + 2x
\end{equation} we get h'(y) = 0 => h(y) = C, C is a constant.At the end we got the answer
\begin{equation}
𝛗(x,y) = x^2y^2 + 2xy +C
\end{equation}
