If $n=2^kN$, where $N$ is odd, then $$\sum_{d\mid n}(-1)^{n/d}\phi(d)=\sum_{d\mid 2^{k-1}N}\phi(d)-\sum_{d\mid N}\phi(2^kd)$$
I have no idea how to seperate things inside the left side. In a nornal sum function it would be easy but any attempt to deal with this one getting me to a false solution. the solution should be zero.
For $d$ dividing $n=2^{k-1}N$, then $n/d$ is even, so that's the first part of the right hand side.
For $d$ not dividing $2^{k-1}N$ but still dividing, $2^{k}N$, then $2^k$ divides $d$, and more specifically, $d = 2^kd'$ for some $d'$ dividing $N$. In that case, $n/d$ divides $N$, which is odd, so $n/d$ is odd. That gives you the second part of the right hand side.