= Solution
The <möbius function> is $mu(1)=1$, $mu(n)=0$ if a prime square divides $n$, and $mu(n)=(-1)^r$ when $n$ is a product of $r$ distinct primes. The <Riemann zeta function> is $zeta(s)=\sum_{n\ge1}n^{-s}$ for $\Re s>1$.
Both sides of the proposed identity are multiplicative. At $n=p^a$, the right side is $1$ for $a=0,1$ and $1+\mu(p)=0$ for $a\ge2$, exactly $mu(p^a)^2$. Absolute convergence permits rearrangement, so
$$
\boxed{\sum_n\frac{\mu(n)^2}{n^s}=\sum_d\frac{\mu(d)}{d^{2s}}\sum_m\frac1{m^s}=\frac{\zeta(s)}{\zeta(2s)}.}
$$
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