Prove that the function

The given function is $f(x)=x^{2}-x+1$.

$\therefore f^{\prime}(x)=2 x-1$

Now, $f^{\prime}(x)=0 \Rightarrow x=\frac{1}{2}$.

The point $\frac{1}{2}$ divides the interval $(-1,1)$ into two disjoint intervals i.e., $\left(-1, \frac{1}{2}\right)$ and $\left(\frac{1}{2}, 1\right)$.

Now, in interval $\left(-1, \frac{1}{2}\right), f^{\prime}(x)=2 x-1<0$.

Therefore, $f$ is strictly decreasing in interval $\left(-1, \frac{1}{2}\right)$.

However, in interval $\left(\frac{1}{2}, 1\right), f^{\prime}(x)=2 x-1>0$.

Therefore, $f$ is strictly increasing in interval $\left(\frac{1}{2}, 1\right)$.

Hence, $f$ is neither strictly increasing nor decreasing in interval $(-1,1)$.