((a ⇔ b) ⇒ c) ⇔ (a ⇔ (b ⇒ c)) tautology, contradiction, or neither? ((a ⇔ b) ⇒ c) ⇔ (a ⇔ (b ⇒ c)). Prove whether its a tautology, a contradiction, or neither.
My attempt: 
((a ⇔ b) ⇒ c) ⇔ (a ⇔ (b ⇒ c))
if I take all F:


*

*((F ⇔ F) ⇒ F) ⇔ (F ⇔ (F ⇒ F))

*[(F ⇒ F ∧ F ⇒ F) ⇒ F] ⇔ (F ⇔ T)

*(F ∧ F ⇒ F) ⇔ (F ⇒ T ∧ F ⇒ F)

*(T ⇒ F) ⇔ (T ∧ F)

*F ⇔ F


Then it becomes T
Could someone check if Im going the right direction?
Im kind of stuck here. I know its neither. but I don't how to continue from here.
 A: You have $3$ variables and therefore at most $2^3 = 8$ combination of truth values which you have to check. Before going through all combinations you should try to simplify the expression (to save work further on), but in your case this seems to lead nowhere.
For checking the combinations your notation is a little verbose, truth tables are more manageable. Evaluating everything at once is a bit hard, so split up the expression into more manageable parts by first evaluating the innermost groups of parentheses, than the second innermost, etc. 
You can do this on the fly in your truth table by writing the truth values of the sub-expressions under the relation signs:
Step 1: (Evaluating innermost expressions.)
| a | b | c || ((a <-> b) -> c)  <->  (a <-> (b -> c)) |
|---|---|---||-----------------------------------------|
| T | T | T ||      T                           T      |
|    ...    ||                   ...                   |

Step 2: (Evaluating second innermost parentheses.)
| a | b | c || ((a <-> b) -> c)  <->  (a <-> (b -> c)) |
|---|---|---||-----------------------------------------|
| T | T | T ||      T     T               T     T      |
|    ...    ||                   ...                   |

Step 3: 
| a | b | c || ((a <-> b) -> c)  <->  (a <-> (b -> c)) |
|---|---|---||-----------------------------------------|
| T | T | T ||      T     T       T       T     T      |
|    ...    ||                   ...                   |

As pointed out in the comments the expression is false in the case $a = F$, $c = T$ (and $b$ either $T$ or $F$).
A: The comments point out that if $a = F$, $c = T$ (and regardless of $b$) the expression is true. It turns out that the expression you wrote has the same truth-value as $a \vee \neg c$.
Mathematica code: 
LogicalExpand[Equivalent[Implies[Equivalent[a, b], c], Equivalent[a, Implies[b, c]]]]

