Is $|9x-1|^3$ differentiable at $1/9$? I think it is, but I'm stuck at showing it. I know that it is equal to:
$$|9x - 1|  \cdot |9x - 1|^2.$$ 
$|9x - 1|$ is not differentiable at $1/9$, and I am pretty sure that $|9x - 1| ^2$ is differentiable there.

We can look separately at the limits of $\frac{f(x)-f(1/9)}{x-1/9}$ (for $x\neq 1/9$) on both sides of $1/9$: the goal is to see if they exist, and (if so) if they are the same. If this is the case, then $f$ is differentiable at $1/9$, and $f^\prime(1/9)$ equals this common limit.
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Start by factorizing $(9x-1)^3$ to make a factor $x-\frac{1}{9}$ appear:

$$(9x-1)^3 = 9\left(x-\frac{1}{9}\right)(9x-1)^2$$



*

*For $x > \frac{1}{9}$, $\lvert 9x-1\rvert^3 = (9x-1)^3$, so that

 $$\frac{f(x) - f(\frac{1}{9})}{x-\frac{1}{9}} = \frac{(9x-1)^3-0}{x-\frac{1}{9}} = 9(9x-1)^2 \xrightarrow[x\to1/9^+]{} 0$$


*For $x < \frac{1}{9}$, $\lvert 9x-1\rvert^3 = (1-9x)^3 = -(9x-1)^3$, so that

 $$\frac{f(x) - f(\frac{1}{9})}{x-\frac{1}{9}} = -\frac{(9x-1)^3-0}{x-\frac{1}{9}} = -9(9x-1)^2 \xrightarrow[x\to1/9^-]{} 0$$

Therefore, the limits on both sides exist and coincide: $f^\prime(1/9)=\lim_{x\to1/9}\frac{f(x) - f(\frac{1}{9})}{x-\frac{1}{9}}$ thus exists.
