It's a cinch to prove it with truth tables, and murder to prove it by 
argument forms.  Since all the argument forms are proved by truth tables,
your teacher should allow a truth table proof.

[A ≡ (B ≡ C)] ≡ [(A ≡ B) ≡ C]
---------------------------
 T ≡ (T ≡ T)  ≡  (T ≡ T) ≡ T
 T ≡ (T ≡ F)  ≡  (T ≡ T) ≡ F
 T ≡ (F ≡ T)  ≡  (T ≡ F) ≡ T
 T ≡ (F ≡ F)  ≡  (T ≡ F) ≡ F
 F ≡ (T ≡ T)  ≡  (F ≡ T) ≡ T
 F ≡ (T ≡ F)  ≡  (F ≡ T) ≡ F
 F ≡ (F ≡ T)  ≡  (F ≡ F) ≡ T
 F ≡ (F ≡ F)  ≡  (F ≡ F) ≡ F

Do equivalence inside the parentheses
If the values are the same on both sides
of the ≡, put T, otherwise put F.
Then erase what you used to get what
you just put down.

[A ≡ (B ≡ C)] ≡ [(A ≡ B) ≡ C]
-----------------------------
[T ≡    T   ] ≡ [   T    ≡ T]
[T ≡    F   ] ≡ [   T    ≡ F]
[T ≡    F   ] ≡ [   F    ≡ T]
[T ≡    T   ] ≡ [   F    ≡ F]
[F ≡    T   ] ≡ [   F    ≡ T]
[F ≡    F   ] ≡ [   F    ≡ F]
[F ≡    F   ] ≡ [   T    ≡ T]
[F ≡    T   ] ≡ [   T    ≡ F]

Now do equivalence inside the brackets
As before, if the values are the same 
on both sides of the ≡, put T, otherwise 
put F. Then erase what you used to get 
what you just put down.
 

[A ≡ (B ≡ C)] ≡ [(A ≡ B) ≡ C]
-----------------------------
   T          ≡     T    
   F          ≡     F    
   F          ≡     F    
   T          ≡     T    
   F          ≡     F    
   T          ≡     T    
   T          ≡     T    
   F          ≡     F    

Now do the final equivalence inside the 
brackets As before, if the values are the 
same on both sides of the ≡, put T, otherwise 
put F. Then erase what you used to get 
what you just put down.

[A ≡ (B ≡ C)] ≡ [(A ≡ B) ≡ C]
-----------------------------
              T         
              T         
              T         
              T         
              T         
              T         
              T         
              T

Since we end up with all T's, that proves
that  [A ≡ (B ≡ C)] ≡ [(A ≡ B) ≡ C] is a
tautology.        

Edwin