I tried this way: Taking 'UE' together and remaining 'PZZL'. So, as 'Z' is repeating, I divide it by $2!$. My answer: $(5!/2!)\cdot 2! = 120$. Is it right? And why do we divide by its factorial if some letters are repeating?
Take 'UE' together and 'PZZLE' is remaining
The $E$ should be together with $UE$, so in fact you have something like $(UE)PZZL$ or $(EU)PZZL$.
So, as 'Z' is repeating I divide it by 2!
That's correct. If we call the two letters $Z_1$ and $Z_2$, there are $2!$ ways to arrange the $Z$'s, but there is only one distinguishable combination: you cannot tell $Z_1$ and $Z_2$ apart.
My answer :- (5!/2!)*2! = 120. Is it right?
Yes. There are $5!$ ways to arrange the $5$ objects $UE$, $P$, $Z$, $Z$ and $L$. $U$ and $E$ are interchangeable so we can multiply by $2!$, but as earlier said the two $Z$'s are not, so we divide by $2!$.