$M=xy^3+y $

$ N=2(x^2y^2+x+y^4) $

$\frac{\partial{M}}{\partial{y}}=3xy^2+1$

$\frac{\partial{N}}{\partial{x}}=2(2xy^2+1)$

$\frac{\partial{M}}{\partial{y}}\neq \frac{\partial{N}}{\partial{x}}$

The given Equation is not an exact differential equation.

$\frac{\frac{\partial{N}}{\partial{x}}-\frac{\partial{M}}{\partial{y}}}{M} =\frac{2(2xy^2+1)-(3xy^2+1)}{xy^3+y} $

$=\frac{(4xy^2+4)-3xy^2-1}{xy^3+y}$

$=\frac{xy^2+1}{xy^3+y} =\frac{xy^2+1}{(xy^2+1)y} =\frac{1}{y}=f(y)$

$ I.F=e^{\int \frac{1}{y}dy}=e^{\log y}=y$

Multiplying the given equation by the integrating factor, we get

$(xy^3+y)y+2(x^2y^2+x+y^4)y=0$

$(xy^4+y^2)dx+2(x^2y^3+xy+y^5)dy=0$

$\frac{\partial{M}}{\partial{y}}=4xy^3+2y$

$\frac{\partial{N}}{\partial{x}}=2(2xy^3+y)=4xy^3+2y$$

The given Equation is an exact differential equation

Solution is given by

$\int Mdx+\int (Terms\ in\ N\ not\ containing\ x)dy=0$

$y=constant\\ \int (xy^4+y^2)dx-\int 2(y^5)dy=c$

$\frac{x^2y^4}{2}+xy^2-\frac{2y^6}{6}=c$

$\frac{x^2y^4}{2}+xy^2-\frac{y^6}{3}=c $