is $|x|^3$ differentiable at [-1, 2] I first defined the function for $x$>0 and $x$&lt;0.

For $x$>0 the function is $x^3$

For $x$&lt;0 function is =-$x^3$

I differentiated it and both cases turned out to be 0.

so left hand limit=RHL=0.

So the function must be differentiable.

am I correct?
Yes, you have
$$\lim_{h \to 0^+} \frac{|h|^3-|0|^3}{h}=\lim_{h \to 0^+} \frac{h^3}{h}=\lim_{h \to 0^+} h^2 =0$$
And
$$\lim_{h \to 0^-} \frac{|h|^3-|0|^3}{h}=\lim_{h \to 0^-}\frac{-h^3}{h}=\lim_{h \to 0^+} -h^2 =0$$
So $|x|^3$ is differentiable at $x=0$ and its derivative is $0$.

In any other $x \in [-1,2] \setminus \{0\}$ you can use the usual derivative rules since there aren't problems with the derivative.
Edit:
If there's function $g(x)=|f(x)|\\$
$g'(x) = sgn(f(x)).f'(x)$