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

$N=\frac{1}{4}(x+xy^2) $

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

$\frac{\partial{N}}{\partial{x}}=\frac{1}{4}(1+y^2) $

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

The given Equation is not an exact differential equation.

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

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

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

$= \frac{3}{x}=f(x) // I.F=e^{\int \frac{3}{x}}dy=e^{3\log x}=x^3$

Multiplying the given equation by the integrating factor, we get

$\left(y+\frac{1}{3}y^3+\frac{1}{2}x^2\right)x^3dx+\left( \frac{1}{4}(x+xy^2)x^3\right)dy=0 $

$\left(yx^3+\frac{1}{3}y^3x^3+\frac{1}{2}x^5 \right)dx+\frac{1}{4}(x^4+x^4y^2)dy=0$

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

$\frac{\partial N}{\partial x}=x^3+y^2x^3 $

$\therefore$ The given Equation is an exact differential equation

Solution is given by

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

y=constant

$ \int\left( yx^3+\frac{1}{3}y^3x^3+\frac{1}{2}x^5\right)dx-\int 0dy=c $

$\frac{x^4y}{4}+\frac{1}{12}y^3x^4+\frac{1}{12}x^6=c$