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:

Given: $$5f(x) + 4f(1/x) = x^2 - 2$$ --- (Eq. 1)

Replace $$x$$ with $$1/x$$: $$5f(1/x) + 4f(x) = \frac{1}{x^2} - 2$$ --- (Eq. 2)

Solve for $$f(x)$$: Multiply (Eq. 1) by 5 and (Eq. 2) by 4, then subtract:

$$25f(x) - 16f(x) = 5(x^2 - 2) - 4(\frac{1}{x^2} - 2)$$

$$9f(x) = 5x^2 - 10 - \frac{4}{x^2} + 8 = 5x^2 - \frac{4}{x^2} - 2$$

Find $$y$$: $$y = 9x^2 f(x) = x^2(5x^2 - \frac{4}{x^2} - 2) = 5x^4 - 2x^2 - 4$$.

Find $$y'$$ for increasing condition: $$y' = 20x^3 - 4x > 0$$.

$$4x(5x^2 - 1) > 0 \implies 4x(\sqrt{5}x - 1)(\sqrt{5}x + 1) > 0$$.

Interval Analysis: Using the number line, the expression is positive in:

$$(-\frac{1}{\sqrt{5}}, 0) \cup (\frac{1}{\sqrt{5}}, \infty)$$