Prove that the function $f$ given by $f(x)=x^{2}-x+1$ is neither strictly increasing nor decreasing on $(-1,1).$

The given function is $f(x)=x^{2}-x+1$.
$\therefore f^{\prime}(x)=2x-1$.
Now,$f^{\prime}(x)=0 \Rightarrow 2x-1=0 \Rightarrow x=\frac{1}{2}$.
The point $x=\frac{1}{2}$ divides the interval $(-1,1)$ into two disjoint intervals,i.e.,$(-1, \frac{1}{2})$ and $(\frac{1}{2}, 1)$.
In the interval $(-1, \frac{1}{2})$,choose a test point $x=0$. Then $f^{\prime}(0)=2(0)-1=-1 < 0$.
Therefore,$f$ is strictly decreasing in the interval $(-1, \frac{1}{2})$.
In the interval $(\frac{1}{2}, 1)$,choose a test point $x=\frac{3}{4}$. Then $f^{\prime}(\frac{3}{4})=2(\frac{3}{4})-1=\frac{3}{2}-1=\frac{1}{2} > 0$.
Therefore,$f$ is strictly increasing in the interval $(\frac{1}{2}, 1)$.
Since the function is strictly decreasing on $(-1, \frac{1}{2})$ and strictly increasing on $(\frac{1}{2}, 1)$,it is neither strictly increasing nor strictly decreasing on the entire interval $(-1, 1)$.