$f\left(x\right)=x^2-mx+1$ and $\mathrm{\alpha }\le \mathrm{\beta }$ are the roots of $f\left(x\right)=0.$
Now, $\mathrm{\alpha }<\mathrm{\beta }<1$ implies that $f\left(1\right)$ and the coefficients of $x^2$ have the same sign.

This gives $f\left(1\right)>0$
$\Rightarrow 1-m+1>0\Rightarrow m<2$
Also, the discriminant is $m^2-4\ge 0.$

So $m\le -2$ or $m\ge +2.$
Taking intersection, we get,

 $m\le -2.$
Also, note that if $m=-2,$ the roots are $-1,\mathrm{}-1.$