Reworking $\sum_{n \leq x} \frac{1}{n^s}$, where $n$ is relatively prime to some fixed $k$ For a fixed integer $k \geq 1$ and real $s>0$ I want to rework the partial sums
$$\sum_{\substack{
  n \leq x \\
  \text{gcd}(k,n) = 1
}}
\frac{1}{n^s}$$ 
in such a way that I can find an asymptotic formula for them.  I've tried to express the $\text{gcd}(k,n) = 1$ condition in various ways involving $\mu$; for example,
$$\sum_{\substack{
  n \leq x \\
  \text{gcd}(k,n) = 1
}}
\frac{1}{n^s} = \sum_{n\leq x}\frac{1}{n^s}\sum_{d | \text{gcd}(k,n)} \mu(d),$$
since $\sum_{d | \text{gcd}(k,n)} \mu(d)$ is equal to 1 if $k$ and $n$ are relatively prime, and 0 if they aren't. But I can't see where to go from here; the difficulty seems to be that $k$ isn't an index of the summation. 
(This is exercise 3.12 in Apostol's Introduction to Analytic Number Theory.)
 A: What follows is too much material for a comment. I would like to point
out that we can also get the aymptotics from the Mellin-Perron formula
where  the difficulty lies  in estimating  a remainder  integral which
does not converge absolutely.

Let the prime factorization of $k$ be given by
$$k = \prod_q q^v.$$
Then we have by inspection that the Euler product of
$$L(t) = \sum_{n\ge 1, \; (k,n) = 1} \frac{1}{n^{s+t}}$$
is given by
$$L(t) = \prod_p \frac{1}{1-p^{-s-t}} \prod_q (1-q^{-s-t})
= \zeta(s+t) \prod_q (1-q^{-s-t}).$$
Now recall the Mellin-Perron formula which says that
$$\sum_{n=1}^x \lambda_n =
\frac{1}{2} \lambda_x
+ \frac{1}{2\pi i} 
\int_{c-i\infty}^{c+i\infty} L(t) x^t \frac{dt}{t}
\quad\text{where}\quad
L(t) = \sum_{n\ge 1} \frac{\lambda_n}{n^t}$$
with $c$ in the half-plane of convergence of $L(t)$
i.e. $\Re(s+t)>1$  or $t>1-s.$ Intersect  this with the  half-plane of
convergence  of  the Heaviside  step  function  which  is $\langle  0,
\infty\rangle$ to find that $c=1/2$ is admissible.
Here $\lambda_n = [[(k,n) = 1]] \times \frac{1}{n^s}.$

Observe that when  $s>0$ there  is a pole  at $t =  1 -  s$. 
Suppose $s$  is an integer. There are two cases, $s=1$ or $s>1.$ 

First case, $s=1$. Then we have a double pole at $t$ zero and
$$\mathrm{Res}_{t=0} \left( L(t) \frac{x^t}{t} \right) = 
(\gamma + \log x) \prod_q (1 - q^{-1}) = 
\frac{\varphi(k)}{k} (\gamma + \log x)$$
and we immediately have
$$\sum_{n=1}^x \lambda_n =
\frac{\varphi(k)}{k} (\gamma + \log x)
+ \frac{1}{2} [[(k, x)=1]] \frac{1}{x^s}
+ \frac{1}{2\pi i} 
\int_{-1/2-i\infty}^{-1/2+i\infty} L(t) x^t \frac{dt}{t}.$$

Second case, $s>1.$ Now the pole at $t$ zero becomes a simple pole
and
$$\mathrm{Res}_{t=0} \left( L(t) \frac{x^t}{t} \right) = 
\zeta(s) \prod_q (1 - q^{-s})$$
giving for the sum the expansion
$$\sum_{n=1}^x \lambda_n = 
\zeta(s) \prod_q (1 - q^{-s})
+ \frac{1}{2} [[(k, x)=1]] \frac{1}{x^s}
+ \frac{1}{2\pi i} 
\int_{-1/2-i\infty}^{-1/2+i\infty} L(t) x^t \frac{dt}{t}.$$
Remark.  Obviously  the difficulty  here  lies  in estimating  the
remainder  integrals. The  term  $\zeta(s+t)$ is  $O(1)$  on the  line
$\Re(t) = -1/2$  when $s>1$ but does not  converge absolutely and when
$s=1$ it is right on the critical line.

Addendum Mon Nov 24 23:34:28 CET 2014. Here is an approach that is
more in line with what  a number theory textbook exercise would likely
ask for.

As was  pointed out  in the  first response we  can use  the following
straightforward simplification:
$$\sum_{n=1, \; (k,n)=1}^x \frac{1}{n^s}
= \sum_{n=1}^x \frac{1}{n^s} \sum_{d|(k,n)} \mu(d)
\\ = 
\sum_{d|k} \mu(d) \sum_{g=1}^{\lfloor x/d \rfloor} \frac{1}{(gd)^s}
= \sum_{d|k} \frac{\mu(d)}{d^s} 
\sum_{g=1}^{\lfloor x/d \rfloor} \frac{1}{g^s}.$$
We once more have two cases.
 First case, $s=1.$
We get from the asymptotic $H_n \sim \log n + \gamma$ the value
$$\sum_{d|k} \frac{\mu(d)}{d} (\log(x/d) + \gamma)
= \log x \prod_q \left(1-\frac{1}{q}\right)
+ \sum_{d|k} \frac{\mu(d)}{d} (\gamma - \log d).$$
Therefore the dominant term plus the next one are given by
$$\frac{\varphi(k)}{k} \log x 
+ \sum_{d|k} \frac{\mu(d)}{d} (\gamma - \log d)$$
where we have determined the constant that does not depend on $x.$
Interestingly enough  the constant from Mellin-Perron  was not correct
which  indicates  that there  is  a  contribution  from the  remainder
integral that we were not able to evaluate.
The next term in the asymptotics of $H_n$ is $\frac{1}{2n}$ and the error
in going from $\log\lfloor w\rfloor$ to $\log w$ is of order $1/w$ which 
indicates that the next term is $\mathcal{O}(1/x)$ as noted by the first 
responder.
 Second case, $s>1.$
The dominant term here is $\zeta(s)$ and we obtain
$$\zeta(s) \sum_{d|k} \frac{\mu(d)}{d^s}
= \zeta(s) \prod_q \left(1-\frac{1}{q^s}\right)$$
which matches the asymptotic from Mellin-Perron. 
A: You have the right start. You might have the idea by now that your initial goal will be to reverse the order of summation. I will let $(k,n)$ denote $\gcd(k,n)$.
Claim: $\displaystyle \sum_{\substack{n \leq x \\ (n,k) = 1}} f(n) = \sum_{d \mid k} \mu(d) \sum_{l \leq x/d} f(ld)$
Proof: As you've noticed,
$$ \sum_{\substack{n \leq x \\ (n,k) = 1}} f(n) = \sum_{n \leq x} \sum_{d \mid (n,k)} \mu(d) f(n).$$
Now we want to reverse the order of summation. For each divisor $d$ of $k$, we sum over multiples of $d$ that are less than or equal to $x$. So thinking of $n = ld$, we want to include only those $l$ such that $ld \leq x$, or rather $l \leq x/d$. Putting these together allows us to complete the reversal, getting
$$ \sum_{d \mid k} \mu(d) \sum_{l \leq x/d} f(ld),$$
and proving the claim. $\diamondsuit$
For us, this means that 
$$\sum_{\substack{n \leq x \\ (n,k) = 1}} \frac{1}{n^s} = \sum_{d \mid k} \frac{\mu(d)}{d^s} \sum_{l \leq x/d} \frac{1}{l^s}.$$
Now the analysis is somewhat straightforward. When $s = 1$, you can use your knowledge of $\displaystyle \sum_{n \leq y} \frac{1}{n}$ to get the desired asymptotic
$$ \frac{\varphi(k)}{k}(\log x + \gamma) + C_k + O\left( \tfrac{1}{x} \right),$$
where $\gamma$ is the Euler-Mascheroni constant and $C_k$ is a constant depending on $k$ which can be given explicitly.
When $s \neq 1$, then you use your knowledge of $\displaystyle \sum_{n \leq y} \frac{1}{n^s}$ to produce a similar asymptotic.
