Prove that $x^2|\sin(\frac1x)|$ is absolutely continuous on $[0,1]$

Could you help me prove that $$f = x^2|\sin(\frac1x)|$$ is absolutely continuous on $$[0,1]$$?

I tried to prove this with this: f has a derivative f′ almost everywhere, the derivative is Lebesgue integrable, and $$f(x)=f(a)+\int _{a}^{x}f'(t)\,dt$$ for all x on [a,b]. The derivative of the function is: $$2x \mid (\sin(1/x)) \mid - \sin(\frac2x)/(2 \mid \sin\frac1x\mid )$$. But I am not sure how to proceed, since there are many pointswhere $$f$$ is not differentiable. Thank you in advance!

It suffices to prove that $$g(x) = x^2 \sin(\frac1x)$$ is absolutely continuous on $$[0, 1]$$, because that implies that $$f = |g|$$ is absolutely continuous as well.
$$g$$ is differentiable on $$[0, 1]$$ with a bounded derivative: $$|g'(x)| = |2x \sin(\frac1x) - \cos(\frac1x) | \le 3$$ for $$0 < x \le 1$$, and $$g'(0) = 0$$.
Therefore $$g$$ is Lipschitz continuous. On a compact interval, Lipschitz continuity implies absolute continuity.
Hints: It is clear that $$f'$$ exists at al but countably many points and $$|f'| \leq 3$$ a.e. Let $$g(x)=\int_0^{x}f'(t)dt$$. Then $$g$$ is absolutely continuous on $$[0,1]$$ and $$g'=f'$$ a.e.. Now consider $$[a,1]$$ where $$0. On this interval there are only a finite number of points where $$f$$ is not differentiable. These points give a partition on $$[a,1]$$. On each subinterval of this partition $$f-g$$ is a constant. By continuity of $$f$$ and $$g$$ conclude that $$f-g$$ is a constant on $$[a,1]$$ for each $$a$$. Now use continuity again to see that this constant does not depend on $$a$$. This proves that there is a constant $$c$$ with $$f(x)=c+\int_0^{x}f'(t)dt$$ for $$0.