$\text{ If the function }f\left(x\right)=\frac{x-1}{c-x^2+1}\text{ does not take any value in the interval }\left[-1,-\frac{1}{3}\right]\text{ , }$then the least integral value that c can not attain is equal to

$\begin{array}{l}\text{ Let }y=f\left(x\right)=\frac{x-1}{c-x^2+1}\\ \text{ Take }y=-t,\text{ where }t\in \left[\frac{1}{3},1\right]\\ \therefore -t=\frac{x-1}{c-x^2+1}\\ \Rightarrow x^2-c-1=\frac{x-1}{t}\Rightarrow x^2-\frac{1}{t}x+\frac{1}{t}-c-1=0\end{array}$

$\text{ As }-t\in \left[-1,-\frac{1}{3}\right]\text{, hence the above must not possess real solution }\left(\sin \text{ce }it is given \right)$

$\begin{array}{l}\therefore {\left(\frac{1}{t}\right)}^2-4\left(\frac{1}{t}-c-1\right)<0\Rightarrow \frac{1}{t^2}-\frac{4}{t}+4\le -4c\\ \Rightarrow c<-\frac{1}{4}{\left(\frac{1}{t}-2\right)}^2\\ \text{ Now, }\frac{1}{3}\le t\le 1\Rightarrow 3\ge \frac{1}{t}\ge 1\Rightarrow 1\ge \frac{1}{t}-2\ge -1\Rightarrow 0\le {\left(\frac{1}{t}-2\right)}^2\le 1\Rightarrow 0\ge -\frac{1}{4}{\left(\frac{1}{t}-2\right)}^2\ge \frac{-1}{4}\\ \text{ Hence, }c\in \left(-\mathrm{\infty },-\frac{1}{4}\right)\end{array}$