How to find all positive integers n: n=\phi(n)+\mu(n)+\tau(n)?
I assume your 3 functions are:
$\phi(n)$ the number of positive integers less than n that are prime to n
$\mu(n)$ the Moebius mu function - 1 for square free n with an even number of prime factors, -1 for square free n otherwise and 0 otherwise
$\tau(n)$ the number of positive divisors of n
Here's the answer to your question. I verify the easy direction, but leave to you the proof that these are the only values that are fixed points of f. If you need help on this direction, post again.
Let $f(n)=\phi(n)+\mu(n)+\tau(n)$. Then $f(n)=n$ if and only if $n$ is prime or $n=9$.
$f(9)=6+0+3=9$ and if $n$ is prime, $f(n)=n-1-1+2=n$