jmc

We expand the left-hand side and then subtract $9$ from both sides, giving $$\frac{x^2-9(4x^2-4x+1)}{4x^2-4x+1}=\frac{-35x^2+36x-9}{4x^2-4x+1}\le 0$$or $\frac{35x^2-36x+9}{4x^2-4x+1}\ge 0.$ Factoring top and bottom gives $$\frac{(5x-3)(7x-3)}{(2x-1{)}^2}\ge 0.$$Letting $f(x)$ denote the left-hand side, we produce a sign table as follows: \begin{array}{c|ccc|c} &$5x-3$ &$7x-3$ &$(2x-1)^2$ &$f(x)$ \\ \hline$x<\frac{3}{7}$ &$-%%DISP_0%%amp;$-%%DISP_0%%amp;$+%%DISP_0%%amp;$+$\\ [.1cm]$\frac{3}{7}<x<\frac{1}{2}$ &$-%%DISP_0%%amp;$+%%DISP_0%%amp;$+%%DISP_0%%amp;$-$\\ [.1cm]$\frac{1}{2}<x<\frac{3}{5}$ &$-%%DISP_0%%amp;$+%%DISP_0%%amp;$+%%DISP_0%%amp;$-$\\ [.1cm]$x>\frac{3}{5}$ &$+%%DISP_0%%amp;$+%%DISP_0%%amp;$+%%DISP_0%%amp;$+$\\ [.1cm]\end{array}We see that $f(x)>0$ when $x<\frac{3}{7}$ or $x>\frac{3}{5}.$ Also, $f(x)=0$ when $x=\frac{3}{7}$ or $x=\frac{3}{5}.$ Hence, the solution set to $f(x)\ge 0$ is $$x\in \boxed{(-\infty ,\frac{3}{7}]\cup [\frac{3}{5},\infty )}.$$