Answer
Prove both inclusions.
Work Step by Step
We have to prove that
$$(A-B)-C=(A-C)-(B-C),$$
where $A$, $B$ and $C$ are sets.
First we will prove that $(A-B)-C\subseteq (A-C)-(B-C).$
Let's take an arbitrary element $x\in (A-B)-C$. This means that $x\in A-B$ and $x\not\in C$, which can be further written as
$$x\in A\text{ and }x\not\in B\text{ and }x\not\in C.$$
Because $x\in A\text{ and }x\not\in C$, it means $x\in A-C$.
Because $x\not\in B$, it means $x\not \in B-C$ because $B-C$ contains a subset of $B$.
Now because $x\in A-C$ and $x\not \in B-C$ it follows that
$$(A-B)-C\subseteq (A-C)-(B-C).\tag1$$
Then we will prove that $(A-C)-(B-C)\subseteq (A-B)-C.$
Let's take an arbitrary element $x\in (A-C)-(B-C)$. This means that $x\in A-C$ and $x\not\in B-C$, which can be further written as
$$x\in A\text{ and }x\not\in C\text{ and }x\not\in B.$$
Because $x\in A\text{ and }x\not\in B$, it means $x\in A-B$.
Because $x\in A-B$ and $x\not\in C$, it means $x\in (A-B)-C$.
We got:
$$(A-C)-(B-C)\subseteq (A-B)-C.\tag2$$
From $(1)$ and $(2)$ we get:
$$(A-B)-C=(A-C)-(B-C).$$
