Question: Solve the given equation $(2xy^2+2y)+(2x^2y+2x)y'=0$
\begin{align}
M(x,y)=2xy^2+2y &\implies M_y(x,y)=4xy+2\notag\\
N(x,y)=2x^2y+2x &\implies N_x(x,y)=4xy+2\notag
\end{align}
Since $M_y=N_x$, so we know that the given differential equation is exact, that is, there exist $\varphi(x,y)$ such that $\varphi_x=M$ and $\varphi_y=N$.
\begin{align}
\varphi_x(x,y) &= 2xy^2+2y \implies \varphi(x,y) = \int 2xy^2+2y dx = x^2y^2+2xy+f(y)\notag\\
\varphi_y(x,y) &= 2x^2y+2x+f'(y) \implies f'(y)=0 \implies f(y)=C\notag
\end{align}
So
$$\varphi(x,y)=x^2y^2+2xy+C$$
$$x^2y^2+2xy=C$$
