I would like to prove this using Euler products:
$$\frac {\zeta(s)}{\zeta(2s)} = \sum_{n=1}^{\infty}\frac {\lvert \mu(n) \rvert}{n^s}$$
I have gotten here, but don't know if this is a correct approach.
$$\frac {\zeta(s)}{\zeta(2s)} = \prod_{p\text{ runs over primes}}^{}(1 + \frac 1{p^s})$$
I appreciate any advice.
Note that $|\mu(n)|$ is multiplicative. Moreover $$|\mu(n)|=\begin{cases} 1 & \text{if there is no prime $p$ such that }p^2\mid n \\ 0 &\text{otherwise}\end{cases}.$$ Now just expand the Euler product of $\zeta(s)/\zeta(2s)$: $$\prod_p \left( 1+\frac{1}{p^s}\right)=\prod_p \left(\sum_{k=0}^n\frac{|\mu(p^k)|}{p^{ks}}\right) = \sum_{n=1}^\infty \frac{|\mu(n)|}{n^s}.$$