If \( 5f(x) + 4f\!\left( \frac{1}{x} \right) = x^2 - 2 \), \( \forall x \neq 0 \) and \( y = 9x^2 f(x) \), then \( y \) is strictly increasing in:
\( \left( -\frac{1}{\sqrt{5}}, 0 \right) \cup \left( 0, \frac{1}{\sqrt{5}} \right) \)
\( \left( -\frac{1}{\sqrt{5}}, 0 \right) \cup \left( \frac{1}{\sqrt{5}}, \infty \right) \)
\( \left( 0, \frac{1}{\sqrt{5}} \right) \cup \left( \frac{1}{\sqrt{5}}, \infty \right) \)
\( \left( -\infty, -\frac{1}{\sqrt{5}} \right) \cup \left( 0, \frac{1}{\sqrt{5}} \right) \)