Separate variables:

$\frac{dy}{1-y+y^2}=(1+x+x^2)dx.$

Complete the square in the denominator: $y^2-y+1={\left(y-\frac{1}{2}\right)}^2+\frac{3}{4}$. Standard integral:

$\int \frac{dy}{{\left(y-\frac{1}{2}\right)}^2+\frac{3}{4}}=\frac{2}{\sqrt{3}}{\tan }^{-1}\left(\frac{2y-1}{\sqrt{3}}\right).$

Right side integrates to $x+\frac{x^2}{2}+\frac{x^3}{3}$. Thus

$\frac{2}{\sqrt{3}}{\tan }^{-1}\left(\frac{2y-1}{\sqrt{3}}\right)=x+\frac{x^2}{2}+\frac{x^3}{3}+C.$

At $x=0,y=\frac{1}{2}$: LHS $=0$, so $C=0$.

At $x=1$: RHS $=1+\frac{1}{2}+\frac{1}{3}=\frac{11}{6}$. So

${\tan }^{-1}\left(\frac{2y(1)-1}{\sqrt{3}}\right)=\frac{\sqrt{3}}{2}\cdot \frac{11}{6}=\frac{11\sqrt{3}}{12}.$

$\therefore 2y(1)-1=\sqrt{3}\tan \left(\frac{11\sqrt{3}}{12}\right).$