QPaperGen

We have:
\[
f(x) = x^2 - x + 1
\]

Differentiating with respect to \( x \), we get:
\[
f'(x) = 2x - 1
\]

Now,
\[
f'(x) = 0 \Rightarrow 2x - 1 = 0 \Rightarrow x = \frac{1}{2}
\]

So, \( x = \frac{1}{2} \) divides the interval \( (-1, 1) \) into two parts:
\[
(-1, \tfrac{1}{2}) \quad \text{and} \quad (\tfrac{1}{2}, 1)
\]

In \( (-1, \tfrac{1}{2}) \), we have:
\[
f'(x) = 2x - 1 < 0 \Rightarrow f \text{ is strictly decreasing}
\]

In \( (\tfrac{1}{2}, 1) \), we have:
\[
f'(x) = 2x - 1 > 0 \Rightarrow f \text{ is strictly increasing}
\]

\(\textbf{Conclusion:} \)
Since \( f \) is decreasing in one part and increasing in another,
\[\text{the function } f(x) = x^2 - x + 1 \text{ is neither strictly increasing nor strictly decreasing on } (-1, 1).
\]