A quick one

First note that $f(x)=9x^{8}+7x^{6}+5x^{4}+3x^{2}-1$ is an even function, so starting an approximation at x=0 is not wise.

Next notice that the first derivative $f'(x)=72x^{7}+42x^{5}+20x^{3}+6x=x(72x^6{}+42x^{4}+20x^{2}+6)$

Since $(72x^{6}+42x^{4}+20x^{2}+6)$ is always positive we expect that there are only 2 real solutions, one positive and one negative. This expectation comes from a negative slope for all negative values of f(x) and a positive slope for all positive values of f(x).

Using Newtons methond with my TI-84 calculator, starting with x=1 (by typing $1\rightarrow x$ ), I did iterations of the following:

$x-(9x^{8}+7x^{6}+5x^{4}+3x^{2}-1)/(72x^{7}+42x^{5}+20x^{3}+6x)\rightarrow x$

this gives $\{1,0.836,0.687,0.565,0.491,0.470,\textbf{0.469...}\}$