Absolute value of differentiable function Given differentiable function $f:\mathbb{R}\to\mathbb{R}$ find points where $|f|$ is not differentiable. 
I think it's not differentiable in all zeros of $f$ and differentiable everywhere else. Is it true? Does it suffice to show that left and right derivatives are different in these points?

In order to show a function is not differentiable at a point, it does suffice to show that the left and right derivatives are different at that point.  In general that is not an equivalent condition, but in the case of the absolute value of a differentiable function, that is the only thing that can go wrong.
The problem is not just when $f(a) = 0$, but when $f$ crosses the $x$ axis at nonzero slope at $x=a$.  Here is an exhaustive list of cases, given $a\in\mathbb R$:


*

*$f(a)\neq 0$.  You can use continuity to find a neighborhood of $a$ on which $|f(x)|=f(x)$, or on which $|f(x)|=-f(x)$, depending on the sign of $f(a)$.  

*$f(a) = 0$ and $f'(a)\neq 0$.  You can use the definition of the derivative to show the left and right derivatives of $|f|$ at $a$ are unequal.  

*$f(a) = 0$ and $f'(a) = 0$.  You can use the definition of the derivative to show that $|f|$ is differentiable at $a$ with $|f|'(a) = 0$.

