Solutions

Consider the functions $f(x)=|x|$ and $g(x)=x^3$. We have that $f$ is not differentiable at $x=0$ and $g$ takes on all real values. Moreover,

$$f(g(x))=\begin{cases}x^3,&x\ge0,\\-x^3,&\text{otherwise.}\end{cases}$$

So

$$(f\circ g)'(x)=\begin{cases}3x^2,&x\ge0,\\-3x^2,&x<0.\end{cases}$$

For $x=0$,

$$\lim_{h\to0^+}\frac{f(g(h))}{h}=\lim_{h\to0^+}\frac{|h^3|}{h}=\lim_{h\to0^+}h^2=0$$

and

$$\lim_{h\to0^-}\frac{f(g(h))}{h}=\lim_{h\to0^-}\frac{|h^3|}{h}=\lim_{h\to0^-}-h^2=0.$$

Hence, $f\circ g$ is differentiable.