I have to prove that $$\mu(n)^2=\sum_{d^2\mid n}\mu (d).$$
The hint is that I have to show that both sides are multiplicative.
However I don't see why if $n,m$ are coprime, then $$\left(\sum_{d^2\mid n}\mu(d)\right)\left( \sum_{d^2\mid m}\mu(d)\right)=\sum_{d^2\mid nm} \mu(d)\,?$$
Here's a proof of the desired result (not using the hint though):
Write $n = \prod p_i^{e_i}$ for positive $e_i$ so that $$\sum_{d^2\mid n} \mu(d) = \sum_{d \mid m} \mu(d) = \begin{cases}1 & m=1 \\ 0 & m>1,\end{cases}$$ where $m = \prod p_i^{\lfloor{e_i/2}\rfloor}.$
$m=1$ precisely when $\left\lfloor\frac{e_i}2 \right\rfloor = 0 \implies e_i = 1 \text{ for each }i \implies n$ squarefree.
$\mu(n)^2 = |\mu(n)|$ follows exactly the same definition.