Proving relationship between Euler's totient function and Mobius function

Question

I'm trying to prove that $$\phi(n)=\sum_{d|n}\mu(d)\phi(d)$$

My attempt is the following:

Mobius inversion formula tells us that, since $$n=\sum_{d|n}\phi(d)$$

then $$\phi(n)=\sum_{d|n}\mu(d)\frac{n}{d}$$

so $$\phi(n)=\sum_{d|n}\mu(d)\sum_{d'|(n/d)}\phi\left({d'}\right)$$ but I don't know how to procceed. Any hint will be appreciated, thanks in advance.

Answer 1

It occurs to me that, as written, this is false. $$\phi(9)=6\ne -1 = 0-2+1 = \mu(9)\phi(9) + \mu(3)\phi(3) + \mu(1)\phi(1)$$