Solve exponential inequality

$$ \left | x \right |^{2x^2 - 3x + 1} \leq 1\tag1$$

I tried to solve it by rewriting $ 1 $ as $ \left | x \right | ^ 0 $. After this substitution I got the equivalent inequation $ 2x^2 - 3x + 1 \leq 1 $, which led to $ x \in \left [ \frac{1}{2}, 1 \right ] $. I am not sure about whether this way of solving it is correct or not.

(you should mean $2x^2-3x+1\leq \color{red}{0}$) This is not correct. This is not equivalent to $(1)$ because you assume that $|x|\gt 1$. In other words, this is equivalent to $(1)$ only when $|x|\gt 1$. (by the way, there is no $x$ such that $\frac 12\le x\le 1$ and $|x|\gt 1$.)

The solution provided by the book approaches two cases, one with $ \left | x \right | \leq 1 $ and the other one with $ \left | x \right | > 1 $. Now, here is where I encounter difficulties. I know how to solve the second case, but I don't really know what I should do in the first case, where $ \left | x \right | \leq 1 $.

For $|x|=1$ or $x=0$, the inequality $(1)$ holds.

For $0\lt |x|\lt 1$, note that $(1)\iff 2x^2-3x+1\color{red}{\ge}0$.

The answer is that $\color{red}{-1\le x\le\frac 12\quad\text{or}\quad x=1}$.