Where is 1/ln|x| continuous?

Away from the special points. On each of the intervals $(-\infty,-1)$, $(-1,0)$, $(0,1)$, $(1,+\infty)$, the function is built from $|\cdot|$, $\ln$ and $x\mapsto\frac1x$ (the logarithm does not vanish there), so it is continuous.

At $0$. As $x\to0$, $\ln|x|\to-\infty$, hence $g(x)=\frac1{\ln|x|}\to0=g(0)$. So $g$ is continuous at $0$.

At $1$ and $-1$. As $x\to1$, $\ln|x|\to0$, so $|g(x)|=\frac1{|\ln|x||}\to+\infty$. The function is unbounded near $1$ and cannot have limit $g(1)=0$. The same happens at $-1$.

Conclusion. $g$ is continuous on $\mathbb{R}\setminus\{-1,1\}$ and discontinuous at $-1$ and $1$.

Remark. Defining a function "by $0$" at a bad point repairs it only when the limit there really is $0$. Here it works at $0$ but not at $\pm1$, where the blow-up is genuine.