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

Let $y=f\left(x\right)=\frac{x-1}{c-x^2+1}$

Take $y=-t$ , where $t\in \left[\frac{1}{3},1\right]$

$\therefore$  $-t=\frac{x-1}{c-x^2+1}$

$\Rightarrow$  $x^2-c-1=\frac{x-1}{t}\Rightarrow x^2-\frac{1}{t}x+\frac{1}{t}-c-1=0$

As  $-t\in \left[-1,-\frac{1}{3}\right]$ , hence the above must not possess real solution

$\therefore$  ${\left(\frac{1}{t}\right)}^2-4\left(\frac{1}{t}-c-1\right)<0\Rightarrow \frac{1}{t^2}-\frac{4}{t}+4\le -4c$

$\Rightarrow c<-\frac{1}{4}{\left(\frac{1}{t}-2\right)}^2$

Now,   $\frac{1}{3}\le t\le l\Rightarrow 1\le \frac{1}{t}-2\le 1\Rightarrow -\frac{1}{4}\le \frac{1}{4}{\left(\frac{1}{t}-2\right)}^2\le 0$  

Hence, $c\in \left(-\infty ,-\frac{1}{4}\right)$