\sum_{d|n}{\mu(d)g\left(\frac{n}{d}\right)=\sum_{d|n}\mu(d)\sum_{d'|\frac{n}{d}}f(d')=\sum_{d|n}\sum_{d'|\frac{n}{d}}\mu(d)f(d')=\sum_{d'|n}\sum_{d|\frac{n}{d'}}\mu(d)f(d')=\sum_{d'|n}\left(\sum_{d|{\frac{n}{d'}}}{\mu(d)\right)f(d')}=\sum_{d'|n}{\delta\left(\frac{n}{d'}\right)f(d')=1\cdot f(n)=f(n)