[tex]\dfrac{\partial\left(2xy^4+\frac1{x+y^2}\right)}{\partial y}=8xy^3-\dfrac{2y}{(x+y^2)^2}[/tex]
[tex]\dfrac{\partial\left(4x^2y^3+\frac{2y}{x+y^2}\right)}{\partial x}=8xy^3-\dfrac{2y}{(x+y^2)^2}[/tex]
so the ODE is indeed exact and there is a solution of the form [tex]F(x,y)=C[/tex]. We have
[tex]\dfrac{\partial F}{\partial x}=2xy^4+\dfrac1{x+y^2}\implies F(x,y)=x^2y^4+\ln(x+y^2)+f(y)[/tex]
[tex]\dfrac{\partial F}{\partial y}=4x^2y^3+\dfrac{2y}{x+y^2}=4x^2y^3+\dfrac{2y}{x+y^2}+f'(y)[/tex]
[tex]f'(y)=0\implies f(y)=C[/tex]
[tex]\implies F(x,y)=x^2y^3+\ln(x+y^2)=C[/tex]
With [tex]y(1)=2[/tex], we have
[tex]8+\ln9=C[/tex]
so
[tex]\boxed{x^2y^3+\ln(x+y^2)=8+\ln9}[/tex]
