Find exhaustive range of $k$ such that $f(x)=\frac{x-1}{k-x^2}$ never belongs to $\left[-1 \:\: \frac{-1}{3}\right]$ Find exhaustive range of $k$ such that $$f(x)=\frac{x-1}{k-x^2}$$ never belongs to $\left[-1 \:\: \frac{-1}{3}\right]$
My try:
Letting $$y=\frac{x-1}{k-x^2}$$ we get
$$yx^2+x-(1+ky)=0$$ and since $x\in \mathbb{R}$ we have $D >0$ we get
$$4ky^2+4y+1 \gt 0$$
Now the above inequality holds true when $y \in \left( -\infty \: -1\right) \cup \left(\frac{-1}{3} \: \infty\right)$
Any clue here?
 A: $f(x)=\frac{x-1}{k-x^2}$ never belongs to $[-1,-\frac{1}{3}]$
$\iff$ For every $y$ satisfying $-1\le y\le -\frac 13$, there are no $x$ such that $y=\frac{x-1}{k-x^2}$
$\iff$ For every $y$ satisfying $-1\le y\le -\frac 13$, there are no $x$ such that $y(k-x^2)=x-1$ and $k-x^2\not=0$
(If $k-x^2=0$, then $x=1$ implying $k=1$. But then $f(0)=-1$, so $k\not=1$.)
$\iff k\not=1$, and for every $y$ satisfying $-1\le y\le -\frac 13$, there are no $x$ such that $yx^2+x-yk-1=0$
$\iff k\not=1$, and if $-1\le y\le -\frac 13$, then $1^2-4y(-yk-1)\lt 0$
(if $k=0$, then $f(x)=\frac{x-1}{-x^2}$. Its range is $[-\frac 14,\infty).$ So, $k=0$ is sufficient.)
$\iff k\not=1$ and $k=0$ is sufficient, and if $-1\le y\le -\frac 13$, then $g(k):=4k(y+\frac{1}{2k})^2-\frac{1}{k}+1\lt 0$
$\iff k\not=1$ and $k=0$ is sufficient, and (($k\lt 0$ and $-\frac 13\lt -\frac{1}{2k}$ and $g(-\frac 13)\lt 0$) or ($k\lt 0$ and $-\frac{1}{2k}\lt -1$ and $g(-1)\lt 0$) or ($k\lt 0$ and $-1\le -\frac{1}{2k}\le -\frac 13$ and $g(-\frac{1}{2k})\lt 0$) or ($k\gt 0$ and $-\frac 13\lt -\frac{1}{2k}$ and $g(-1)\lt 0$) or ($k\gt 0$ and $-\frac{1}{2k}\lt -1$ and $g(-\frac 13)\lt 0$) or ($k\gt 0$ and $-1\le -\frac{1}{2k}\le -\frac 13$ and $g(-1)\lt 0$ and $g(-\frac 13)\lt 0$))
$\iff k\not=1$ and $k=0$, and ($k\lt 0$ or $0\lt k\lt\frac 12$ or $\frac 12\le k\lt \frac 34$)
$$\iff \color{red}{k\in\left(-\infty,\frac 34\right)}$$

Another way : 
Let us separate it into cases :


*

*Case 1 : If $k=0$, then $f(x)=\frac{1-x}{x^2}$ and $f'(x)=\frac{x-2}{x^3}$, and its range is $[f(2),\infty)$, i.e. $[-\frac 14,\infty)$.

*Case 2 : If $k\in\mathbb [\frac 34,\infty)$, then $f\left(\frac{1-\sqrt{4k-3}}{2}\right)=-1$ with $k-\left(\frac{1-\sqrt{4k-3}}{2}\right)^2\not=0$.

*Case 3 : If $k\lt 0$, then $k-x^2\lt 0$ for every $x$. Then, we have $$f(x)\gt -\frac 13\iff x^2-3x+3-k\gt 0$$ which holds for every $x$ since $D=4k-3\lt 0$. So, $f(x)\gt -\frac 13$ for every $x$.

*Case 4 : If $k\in(0,\frac 34)$, then $f(x)$ is not defined when $x=\pm\sqrt k$. We have $$f'(x)=\frac{(x-(1+\sqrt{1-k}))(x-(1-\sqrt{1-k}))}{(k-x^2)^2}$$ with $-\sqrt k\lt 0\lt 1-\sqrt{1-k}\lt\sqrt k\lt 1\lt 1+\sqrt{1-k}$ and $$\lim_{x\to\pm\infty}f(x)=0,\quad \lim_{x\to {-\sqrt k}^{\pm}}f(x)=\mp\infty,\quad \lim_{x\to{\sqrt k}^{\pm}}f(x)=\pm\infty$$$$f(1\pm\sqrt{1-k})=\frac{-1\pm\sqrt{1-k}}{2k}\lt 0,\quad f(0)=-\frac 1k\lt 0,\quad f(1)=0$$So, considering the graph of $y=f(x)$, a necessary and sufficient is $$f(1-\sqrt{1-k})\lt -1\qquad\text{and}\qquad f(1+\sqrt{1-k})\gt -\frac 13,$$i.e.$$0\lt k\lt\dfrac 34$$
Therefore, from the four cases, the answer is
$$k\in\left(-\infty,\frac 34\right)$$
