Answer
Sometimes true.
Work Step by Step
It's not always true, because when $b=0$ or $c=0$, the division is undefined.
Let us see if it is ever true.
$LHS=(a\div b)\div c=(a \cdot \displaystyle \frac{1}{b})\div c\qquad $( definition of division)
$=(a \cdot \displaystyle \frac{1}{b})\cdot \displaystyle \frac{1}{c}\qquad $( definition of division)
$=a \cdot( \displaystyle \frac{1}{b}\cdot \displaystyle \frac{1}{c})\qquad $(associative property of multiplication)
$=a\displaystyle \cdot\frac{1}{bc}\qquad $(simplify $\displaystyle \frac{1}{b}\cdot \displaystyle \frac{1}{c}$)
$=\displaystyle \frac{a}{bc}$
$ RHS= a\div(b\div c)=a\div(b\cdot \displaystyle \frac{1}{c})\qquad $( definition of division)
$=a\div \displaystyle \frac{b}{c}\qquad $(simplify $ b\cdot \displaystyle \frac{1}{c}$)
$=a\displaystyle \cdot\frac{1}{\frac{b}{c}}\qquad $(definition of division)
$=a\displaystyle \cdot\frac{c}{b}\qquad $(simplify $\displaystyle \frac{1}{\frac{b}{c}}$)
$=\displaystyle \frac{ac}{b}\qquad $(simplify)
So, RHS =LHS when
$\displaystyle \frac{a}{bc}=\frac{ac}{b}$
We see that when $c=\pm 1,b\neq 0$, then LHS=RHS,
so the statement is sometimes true.
(It is true when $a=0$ as well, $( a$, the numerator, can be zero)
but it was enough to find one instance when the statement is valid
to conclude that it is sometimes true.)
