On differentiating w.r.t. $x$, we get
$f^{\prime}(x)=\frac{\left(x^2+x+1\right)(2 x-1)-\left(x^2-x+1\right)(2 x+1)}{\left(x^2+x+1\right)^2}$
for maximum or minimum, put $f^{\prime}(x)=0$
$\begin{array}{lc}
\Rightarrow & \left(x^2+x+1\right)(2 x-1)-\left(x^2-x+1\right)(2 x+1)=0 \\
\Rightarrow & x^2+x-1-\left(-x^2+x+1\right)=0 \\
\Rightarrow & 2 x^2-2=0 \Rightarrow x= \pm 1
\end{array}$
Now, $f^{\prime}(c)=\frac{2 x^2-2}{\left(x^2+x+1\right)^2}$
Again differentiating, we get
$\begin{aligned}
& \left(x^2+x+1\right)^2(4 x)-\left(2 x^2-2\right) \\
& f^{\prime \prime}(x)=\frac{2\left(x^2+x+1\right)(2 x+1)}{\left(x^2+x+1\right)^4}
\end{aligned}$
at $x=1, f^{\prime \prime}(x)>0$
Therefore it is minimum at $x=1$
Put $x=1$ in equation (i), we get
$f(1)=\frac{1-1+1}{1+1+1}=\frac{1}{3}$
$\therefore$ The minimum value is $\frac{1}{3}$.