How to solve a logarithmic inequality? I mean
 $$\left(\log _{\frac{3 x^2}{2}-\frac{4 x}{3}+\frac{5}{6}}\left(4 x^2+1\right) \log _{\frac{3 x^2}{2}-\frac{4 x}{3}+\frac{5}{6}}\left(1-4 x^2\right)+1\right) \log _{1-16 x^4}\left(\frac{3 x^2}{2}-\frac{4 x}{3}+\frac{5}{6}\right)\geq 1. $$
My (unsuccessful) try was to replace $\log _{1-16 x^4}\left(\frac{3 x^2}{2}-\frac{4 x}{3}+\frac{5}{6}\right) $ by $\frac 1 {\log _{\frac{3 x^2}{2}-\frac{4 x}{3}+\frac{5}{6}}\left(1-16 x^4\right)} $ and to determine the domains of the logarithms. My Mathematica 12.0 is running for hours without any result, solving it.
Let $$f=\frac{3 x^2}{2}-\frac{4 x}{3}+\frac{5}{6},\quad g=4x^2+1,\quad h=1-4x^2$$ 

Then, your inequality is

$$\left(\log _{f}\left(g\right) \log _{f}\left(h\right)+1\right) \log _{gh}\left(f\right)\geq 1, $$
i.e.
$$\left(\log _{f}\left(g\right) \log _{f}\left(h\right)+1\right)\frac{1}{\log _{f}(gh)}\geq 1\tag1$$

By the way, we have to have
$$f\gt 0,f\not=1,g\gt 0,h\gt 0,gh\not=1,$$
i.e.
$$x\in \left(-\frac{1}{2},-\frac 19\right)\cup \left(-\frac 19,0\right)\cup\left(0,\frac{1}{2}\right)\tag2$$

Now, let us separate it into two cases : 

Case 1 : If $\log_f(gh)\gt 0$, then
$$\begin{align}(1)&amp;\iff \log _{f}\left(g\right) \log _{f}\left(h\right)+1\ge \log_f(g)+\log_f(h)\\\\&amp;\iff (\log _{f}\left(g\right)-1)(\log _{f}\left(h\right)-1)\ge 0\tag3\end{align}$$

Here, noting that $gh=1-16x^2\lt 1$ under $(2)$, we have 
$$\log_f(gh)\gt 0\iff f\lt 1\iff x\in\left(-\frac 19,0\right)\cup\left(0,\frac{1}{2}\right)\tag4$$

Also, noting that $f-g=-\frac{1}{6}(5x+1)(3x+1)\lt 0$ under $(4)$, we have $\log_f(g)-1\lt 0$.

So, we have
$$(3)\iff \log _{f}\left(h\right)-1\le 0\iff f-h=\frac 16(3x-1)(11x+1)\le 0$$

So, in this case, we have
$$x\in \bigg[-\frac{1}{11},0\bigg)\cup\bigg(0,\frac 13\bigg]$$

Case 2 : If $\log_f(gh)\lt 0$, then
$$(1)\iff (\log _{f}\left(g\right)-1)(\log _{f}\left(h\right)-1)\le 0\tag5$$

Since $gh\lt 1$, we have
$$\log_f(gh)\lt 0\iff f\gt 1\iff x\in\left(-\frac{1}{2},-\frac 19\right)\tag6$$
Also, noting that $f-h=\frac 16(3x-1)(11x+1)\gt 0$ under $(6)$, we have $\log_f(h)-1\lt 0$.

So, we have
$$(5)\iff \log_f(g)-1\ge 0\iff f-g=-\frac{1}{6}(5x+1)(3x+1)\le 0$$

So, in this case, we have
$$x\in \bigg(-\frac{1}{2},-\frac 13\bigg]\cup\bigg[-\frac 15,-\frac 19\bigg)$$

Therefore, the answer is
$$\color{red}{x\in \bigg(-\frac{1}{2},-\frac 13\bigg]\cup\bigg[-\frac 15,-\frac 19\bigg)\cup\bigg[-\frac{1}{11},0\bigg)\cup\bigg(0,\frac 13\bigg]}$$
Hint: It simplifies to
$$\frac{\ln(4x^2+1)\ln(1-4x^2)}{\ln(1-16x^4)\ln(\frac{3}{2}x^2-\frac{4}{3}x+\frac{5}{6})}+\frac{\ln(\frac{3}{2}x^2-\frac{4}{3}x+\frac{5}{6})}{\ln(1-16x^4)}\geq 1$$
And this is $$\frac{1}{\ln(1-16x^4)}\left(\frac{\ln(4x^2+1)\ln(1-4x^2)}{\ln(\frac{3}{4}x^2-\frac{4}{3}x+\frac{5}{6})}+\ln(\frac{3}{2}x^2-\frac{4}{3}x+\frac{5}{6})\right)\geq 1$$