This solution is different than the others shown. First, consider the simple function $f(x) = |x|$. We want to know $f'(x)$. That's actually pretty simple to do, since $f'(x)$ gives the slope of the tangent line. That is -1 for $x < 0$ and 1 for $x > 0$. At $x = 0$, there is a sharp point in $f(x)$ so $f'(x)$ does not exist. Or, another way to see it, the left hand limit of the difference quotient is -1 and the right hand limit is +1, so the limit itself (the derivative) does not exist. Therefore, a very nice shorthand way of writing the derivative is
$$f'(x) = \frac{x}{|x|}.$$
This is -1 when $x$ is negative, undefined when $x$ is 0, and 1 when $x$ is positive, just as I described above. When you have a more complicated function inside the absolute values, just use the chain rule.
If $f(x) = |x^2 - 2x|$, then
$$f'(x) = \frac{x^2 - 2x}{|x^2 - 2x|} \cdot \frac{d}{dx} (x^2 - 2x) = \frac{x^2 - 2x}{|x^2 - 2x|} (2x - 2) = \frac{2x(x - 1)(x - 2)}{|x^2 - 2x|}.$$
Now, critical points are those where the derivative is 0 or undefined, and the original function is defined. Since the function $f(x)$ is defined for all $x$, find where the derivative is 0 or undefined. Just find the places where the numerator is 0 or the denominator is 0. That's pretty simple here, $x = 0, 1, 2$.