All real values of $k$ in rational function 
All real values of $k$ for which the range of function $\displaystyle f(x) = \frac{x-1}{k-x^2+1}$ does not contain the interval $\displaystyle \left[-1,-\frac{1}{3}\right].$

$\bf{}My\; Try::$ Let $\displaystyle y = \frac{x-1}{k-x^2+1}\Rightarrow ky-x^2y+y=x-1\Rightarrow x^2y+x-ky-y-1=0$
Now for real values of $y\;,$ We must have $x$ real. So Put $\bf{Discriminant\geq 0}$
So we get $1+4y(ky+y+1)\geq0\Rightarrow 4ky^2+4y^2+4y+1\geq0$
So We get $4(k+1)y^2+4y+1\geq0\;$
Now How can I solve after that, Help required, Thanks

As you did, considering the discriminant, we get
$$(4k+4)y^2+4y+1\ge 0\tag1$$
Now, let us separate it into cases.
Case 1 : $k\lt -1$
$$(1)\iff (-4k-4)y^2-4y-1\le 0\iff \frac{1-\sqrt{-k}}{2(-k-1)}\le y\le \frac{1+\sqrt{-k}}{2(-k-1)}$$
So, we need
$$\frac{1+\sqrt{-k}}{2(-k-1)}\lt -1\quad\text{or}\quad \frac{1-\sqrt{-k}}{2(-k-1)}\gt -\frac 13$$
Solving these gives
$$k\lt -1\quad\text{or}\quad -1\lt k\lt -1/4$$
So, this case is sufficient.
Case 2 : $k=-1$ 
Since $(1)\iff y\ge -1/4\gt -1/3$, this is sufficient.
Case 3 : $-1\lt k\lt 0$
$$(1)\iff y\le\frac{-1-\sqrt{-k}}{2(k+1)}\quad\text{or}\quad y\ge \frac{-1+\sqrt{-k}}{2(k+1)}$$
So, we need
$$\frac{-1-\sqrt{-k}}{2(k+1)}\lt -1\quad\text{and}\quad \frac{-1+\sqrt{-k}}{2(k+1)}\gt -\frac 13$$
Solving these gives
$$-1\lt k\lt -\frac 14$$
So, in this case, we have $-1\lt k\lt -1/4$.
Case 4 : $k\ge 0$ 
$(1)$ holds for every $y$, so this case does not satisfy the condition.
Therefore, the answer is $\color{red}{k\lt -1/4}$.
