Are $\frac{a}{\gcd(a,n)}$ and $n$ always relatively prime? 
If $d = \gcd(a,n)$, then $\dfrac ad$ and $n$ must be relatively prime. Prove or
  disprove.

Do I have to show that they need to be relatively prime and then the inverse that they do not need to be relatively prime? I do not know where to start. I think that they do not need to be relatively prime, but I cannot think of an example. 
 A: The simplest counterexample is $n^2$ and $n$.
A: Let $n=6$ and $a=12$. Then $d=\gcd(12,6)=6$. But $\frac ad = \frac{12}6 = 2$.
A: Though the given answers provide some good counterexamples but this question is not totally about giving counterexamples (As you say: Prove or Disprove).
I shall use many variables here, so read with a little care.
CLAIM: If $d=gcd(a,n)$, then both the possibilities are open (i.e.  $\frac{a}{d}$ and $n$ may be coprime or may be not).
Let's see the proof now:
Since $d=gcd(a,n)$ so we can say that $a=dk$ and $n=dl$, where $d,l$ are coprime integers (WHY ??  This is your exercise).
Now, $\frac{a}{d}=k$ and as known, $n=dl$. What we are trying to find is that  $k,dl$ are coprime or not. 
Notice that I have already mentioned that $k,l$ are coprime, but no such conclusion is made regarding k and d. So, if k and d share a factor then $\frac{a}{d}$ and $n$ will not be coprime and if k and d are coprime then $\frac{a}{d}$ and $n$ will also be coprime.
Conclusion:
If $n$ is a factor of $a$ and on dividing $a$ by $n$ there is still something in $a$ which was also in $n$ then $\frac{a}{d}$ and $n$ will not be coprime otherwise they will be coprime.
Examples: The given examples illustrate the conclusion very well.
In @abnry's example: $n$ divides $n^2$  and after that, we are still left with the $n$.
In @zelzy's example: $6$ divides $12$ and after that we are left with $2$ which is a factor of $6$.
A: Let us disprove a slightly weaker statement:

given integers $a, b$, not both zero, $$\text{either } \gcd\left(a, \frac{b}{\gcd(a, b)}\right) = 1 \text{ or }\gcd\left(\frac{a}{\gcd(a, b)}, b\right) = 1$$

As a counterexample, just take $a = 2^{2} \cdot 3 = 12$ and $b = 2 \cdot 3^{2} = 18$.
A: $\gcd(a,n)| a; \gcd(a,n)|n$ so let $a' =\frac a{\gcd(a,n)}$ and $n' = \frac {n}{\gcd(a,n)}$.  So $a = a'*\gcd(a,n)$ and $n = n'*\gcd(a,n)$.  This "separates" $a$ and $n$ into distinct components.  This is always a good place to start.
[  Clearly $\gcd(a',n') = 1$ because if $k|a'$ and $k|n'$ then $k*\gcd(a,n)|a$ and $k*\gcd(a,)|n$ and so $\gcd(a,n)$ is not the greatest common divisor.]
So are $\frac a{\gcd(a,n)}= a'$ and $n = n'\gcd(a,n)$ always relatively prime?  Well, $n'$ and $a'$ have no common factors.  But do $a'$ and $\gcd(a,n)$ have a common factor?  
Well, they might have.  If $a$ had a higher power of factor than $n$ did.  
Example:  Suppose $\gcd(a,n) = k^m*d$, $a = \overline{a}k^{m+s}d $ and $n = n'k^m*d$.  (In other words, $a' = k^s*\overline{a}$.)  Then $\frac {a}{\gcd(a,n)} = k^s\overline{a}$ and $n = n'*k^m*d$.  These are not relatively prime.
Example: $\gcd(a,n) = 4^3*7$ and $a = 27*4^5*7$ and $n = 143*4^3*7^3$.  Then $\frac {a}{\gcd(a,n)} = 27*4^2$ while $n = 143*4^3*7^3$ and they are not relatively prime.
A far easier example will be: $a = n^2$ and $n =n$  Then $\gcd(a,n) = n$ and $\frac a{\gcd(a,n)} = \frac {n^2}{n} = n$.  $n$ and $n$ are, of course, not relatively prime.
A: i am writing again because my first solution has confused some people.  DON'T MIND!!!
$GCD(a,n)=d$ 
by the definition of GCD there is no other common factor in $a,n$
now we would try to prove that if some factor is common between them then it must be in $d$
now let $\frac{a}{d},n$ are not coprime 
$\implies$GCD(\frac{a}{d},n)=x$
$\implies x|\frac{a}{d}$ 
$\implies x|a$
&    $x|n$
$\implies x|GCD(a,n)=d$
we had eliminated every thing common in $a,n$ by dividing $a$ by $d$ from $a$ so $\frac{a}{d},n$ are coprime
A: let us assume that $\frac{a}{d},n$ are not relatively primes 
$\implies n,\frac{a}{d}$has a common factor.let this factor be $x$
$\implies x|a,n$
as $x$ is a factor of $\frac{a}{d}$
$\implies$ $x$ survives
$\implies$ $x$ is not contained in $d$ with that multiplicity by which it can  eliminate it completely from $a$
$x$ must not contained in $n$ if contained in,it must be in $d$
a contradiction!!!
