We know,IF $|f(x)| \geqslant a, \quad ;a>0$
then,
$$f(x) \geqslant a \quad \cup \quad f(x) \leqslant-a$$
$\frac{x^{2}+1}{x}> 6$ & $\frac{x^{2}+1}{x}<-6$
$$\frac{x^{2}-6 x+1}{x}>0..........(1)$$
$$\frac{x^{2}+6 x+1}{x}<0............(2)$$
From (1) ,I can find,
$$x \in(0,3-2 \sqrt{2}) \cup(3+2 \sqrt{2}, \alpha)$$
From (2) ,I can find,
$$x \in(-\alpha,-3-2 \sqrt{2}) \cup(-3+2 \sqrt{2},0)$$
So the domain is :
$$(-\infty,-3-2\sqrt2)\cup(-3+2\sqrt2,0)\cup(0,3-2\sqrt2)\cup(3+2\sqrt2,+\infty)$$