Is the following proof okay?
First let $U(n):=\sum\limits _{d^{2}|n} \mu (n)$. Consider $m,n\in \mathbb{N}$ such that $(m,n)=1$. We have that $$U(m)\cdot U(n)=\sum\limits _{d^{2}|m} \mu (d)\cdot \sum\limits _{e^{2}|n} \mu (e)=\sum\limits _{(de)^{2}|mn} \mu (de)=U(m\cdot n)$$ since $d^{2}\mid m$ and $e^{2}\mid n$ implies that $(de)^{2}\mid mn$. Thus $U(n)$ is multiplicative.
No, if we let $n=p_{1}^{a_{1}}\cdots p_{s}^{a_{s}}$, then $$U(n)=U(p_{1}^{a_{1}}\cdots p_{s}^{a_{s}})=U(p_{1}^{a_{1}})\cdots U(p_{s}^{a_{s}}).$$ Therefore we only need to consider $U(p^{\alpha })$. Note that when $\alpha \leq 1$ then $$U(p^{\alpha})=\sum\limits _{d^{2}|p^{\alpha}} \mu (d)=\mu (1)=1.$$ If $\alpha >1$, then $$U(p^{\alpha})=\sum\limits _{d^{2}|p^{\alpha}} \mu (d)=\mu (a)+ \mu (p)=1-1=0,$$ which we wanted to show.