Let $K^*=K-\{0\}$ be the unit group in the field $K$, and $G$ be a finite subgroup of $K^*$ with $|G|=n$ (Note that $G$ is just known to be a finite Abelian group under mutiplication). $G_d$ is the subset consisting of all the d-order elements in $G$. We need to prove $$n=\Sigma_{d|n,G_d \neq \emptyset}\phi(d)$$
By obesevation, every term in the RHS is represented by the cardinality of all the d-order elements in a cyclic group $\{1,x,x^2,x^3...\}$ generated by some d-order element $x$. In other words, I should prove for any d-order element $b\in G_d$, there exists $i\in\mathbb{N}$ such that $b=x^i$. I should use some property of $K^*$ and consider the elements of the form $x^ib^j$, but I failed to see how to imply the conclusion $b=x^i$.
Any help is highly appreciated!
As you said, your equality is easy to get once you know that $G$ is cyclic. This is actually true : any finite subgroup of the group of the non-zero elements of a field is always cyclic. (this is the case in particular for the $\mathbb{Z}/p\mathbb{Z}^* \cong \mathbb{Z}/(p - 1)\mathbb{Z}$ when $p$ is prime). To prove it, consider the polynomial
$P = \prod_{g \in G} (X - g)$ that has coefficients in $K$. Clearly, $G$ is the zero locus of $P$ and for all $g \in G$, $g^n = 1$. Therefore, $P$ and $X^n - 1$ have the same zero locus. They also have the same degree, the same dominant coefficient and they both only have simple roots (this is trivial for $P$ and $\frac{d}{dX}(X^n - 1) = nX^{n - 1}$ doesn't vanish when $X^n - 1$ does). It implies that $P = X^n - 1$.
Now, write $G \cong \mathbb{Z}/d_1\mathbb{Z} \times \cdots \times \mathbb{Z}/d_k\mathbb{Z}$ with for all $i$, $d_i|d_{i + 1}$ using Kroenecker's theorem. In this case, for any $g \in G$, $g^{d_k} = 1$. Therefore, $X^n - 1 = P|X^{d_k} - 1$ by looking at the roots. It implies that $d_k = n$ thus all the other $d_i$ are $1$ so $G \cong \mathbb{Z}/n\mathbb{Z}$, which proves the wanted result.