Elements in a finite group of order prime

You almost have it, you just need to flip things around a bit:

$\begin{equation*} (x^{-1})^{100} \equiv 1 \quad mod(101) \end{equation*}$

$\begin{equation*} ((x^{-1})^{33})^{3}(x^{-1}) \equiv 1 \quad mod(101) \end{equation*}$

$\begin{equation*} ((x^{-1})^{33})^{3} \equiv x \quad mod(101) \end{equation*}$

So the cube root of $x$ is $(x^{-1})^{33}$.

Hint: Fermat's little theorem also tells you that $\ x^{101}\equiv x \mod(101)\ $, and therefore that $\ x^{101+100n}\equiv x \mod(101)\ $ for any integer $\ n\ $. What happens if you choose $\ n=1\ $?

By the way, the group of units $\ (\mathbb{Z}/p\mathbb{Z})^*$ isn't a field. It's a cyclic group under multiplication, but it's not closed under addition. The entire ring $\ \mathbb{Z}/p\mathbb{Z}\ $ is what constitutes a field.

Consider the map $f: x \mapsto x^3$. Since your group is abelian, it is easy to check that $f$ is a group homomorphism. Now, the order of your group is $\phi(101) = 100$ and so no non-identity element has an order which is either a divisor or a multiple of 3. So, $\text{ker}(f) = \{x \in (\mathbb{Z}/101 \mathbb{Z})^*| x^3 =1\} = \{1\}$ and hence $f$ is injective. Since $f$ is an injection from a finite set to itself, $f$ is also a surjection and therefore, every element has a pre-image via $f$. That is, every element is a cube.