In order to show that equality does not always hold, you should produce specific examples of $A$, $B$ and $C$ where the two sides do not agree.
Note that the left-hand side is equivalent to
$$\begin{align*}
A\setminus (B\setminus C) &= A\setminus (B\cap C^c)\\
&= A\cap (B\cap C^c)^c\\
&= A\cap (B^c\cup C).
\end{align*}$$
That is, the things that are in $A$ and either in $C$ or not in $B$.
On the other hand, the right hand side is equivalen to:
$$\begin{align*}
(A\setminus B)\setminus C &= (A\cap B^c)\setminus C\\
&= (A\cap B^c)\cap C^c\\
&= A\cap B^c\cap C^c
\end{align*}$$
that is, the things that are in $A$, and not in $B$, and not in $C$.
So an easy way to find examples is to find one in which there is an element that is in both $A$ and $C$; then it will be on the left-hand side but not on the right-hand side.
So take $A=\{1\}$, $B=\varnothing$, $C=\{1\}$. Then
$$A\setminus(B\setminus C) = \{1\}\setminus(\varnothing\setminus \{1\}) = \{1\}\setminus\varnothing = \{1\},$$
but
$$(A\setminus B)\setminus C = (\{1\}\setminus\varnothing)\setminus\{1\} = \{1\}\setminus\{1\} = \varnothing.$$
This proves that equality does not always hold.
Note that there are cases where the two expressions are equal; for example, if $C$ is empty. More generally, if $A$ is disjoint from $C$, then $A\subseteq C^c$, so $(A\setminus B)\setminus C = A\setminus B$, and
$$A\setminus (B\setminus C) = A\cap (B^c\cup C) = (A\cap B^c)\cup (A\cap C) = A\cap B^c$$
so the two are equal. This is the only situation where you have equality: if $x\in A\cap C$, then $x\in A\setminus(B\setminus C)$, but $x\notin $A\cap B^c\cap C^c=(A\setminus B)\setminus C$.