The cube of 99 is 970299, which ends with 99. Consider the smallest two-digit number whose cube ends with the original two-digit number.
Let $n$ be the two-digit number whose cube ends with $n$. In other words, $n^3 \equiv n \pmod{100}$. Since $n^3 \equiv n \pmod{100}$, we must have $n^3 \equiv n \pmod{25}$ and $n^3 \equiv n \pmod{4}$. Therefore, we can find a two-digit number whose cube ends with $n$ by finding a two-digit number such that $n^3 \equiv n \pmod{25}$ and $n^3 \equiv n \pmod{4}$.
Using the Chinese Remainder Theorem, we want to find the number $n$ such that $n^3 \equiv n \pmod{100}$. This is equivalent to finding $n$ such that $n^3 \equiv n \pmod{25}$ and $n^3 \equiv n \pmod{4}$.
Consider the integers from $0$ to $24$:
$0, 1, 2, 3, 4, \ldots, 23, 24$.
We test each integer, squaring and multiplying by the integer until powers of integers repeat using the algorithm below.
\begin{align*}
0 &\implies 0^2 \equiv 0 \pmod{25} \equiv 0 \\
1 &\implies 1^2 \equiv 1 \pmod{25} \equiv 1 \\
2 &\implies 2^2 \equiv 4 \pmod{25} \equiv 4 \\
3 &\implies 3^2 \equiv 9 \pmod{25} \equiv 9 \\
4 &\implies 4^2 \equiv 16 \pmod{25} \equiv 16 \\
5 &\implies 5^2 \equiv 0 \pmod{25} \equiv 0 \\
6 &\implies 6^2 \equiv 1 \pmod{25} \equiv 1 \\
7 &\implies 7^2 \equiv 4 \pmod{25} \equiv 4 \\
8 &\implies 8^2 \equiv 9 \pmod{25} \equiv 9 \\
9 &\implies 9^2 \equiv 16 \pmod{25} \equiv 16 \\
10 &\implies 10^2 \equiv 0 \pmod{25} \equiv 0 \\
11 &\implies 11^2 \equiv 1 \pmod{25} \equiv 1 \\
12 &\implies 12^2 \equiv 4 \pmod{25} \equiv 4 \\
13 &\implies 13^2 \equiv 9 \pmod{25} \equiv 9 \\
14 &\implies 14^2 \equiv 16 \pmod{25} \equiv 16 \\
15 &\implies 15^2 \equiv 0 \pmod{25} \equiv 0 \\
16 &\implies 16^2 \equiv 1 \pmod{25} \equiv 1 \\
17 &\implies 17^2 \equiv 4 \pmod{25} \equiv 4 \\
18 &\implies 18^2 \equiv 9 \pmod{25} \equiv 9 \\
19 &\implies 19^2 \equiv 16 \pmod{25} \equiv 16 \\
20 &\implies 20^2 \equiv 0 \pmod{25} \equiv 0 \\
21 &\implies 21^2 \equiv 1 \pmod{25} \equiv 1 \\
22 &\implies 22^2 \equiv 4 \pmod{25} \equiv 4 \\
23 &\implies 23^2 \equiv 9 \pmod{25} \equiv 9 \\
24 &\implies 24^2 \equiv 16 \pmod{25} \equiv 16 \\
\end{align*}
Since $5^2 \equiv 0 \pmod{25}$, $10^2 \equiv 0 \pmod{25}$, $15^2 \equiv 0 \pmod{25}$, and $20^2 \equiv 0 \pmod{25}$, it follows that the set of residues generated by squaring numbers modulo $25$ is $\{0, 1, 4, 9, 16\}$.
We test each integer modulo $4$:
\begin{align*}
0 &\implies 0^3 \equiv 0 \pmod{4} \equiv 0 \\
1 &\implies 1^3 \equiv 1 \pmod{4} \equiv 1 \\
2 &\implies 2^3 \equiv 0 \pmod{4} \equiv 0 \\
3 &\implies 3^3 \equiv 1 \pmod{4} \equiv 1 \\
\end{align*}
Since $2^3 \equiv 0 \pmod{4}$, it follows that the set of residues generated by cubing numbers modulo $4$ is $\{0, 1\}$.
Thus, we need to find an integer that is congruent to $n \pmod{25}$ for $n \in \{0, 1, 4, 9, 16\}$ and is congruent to $n \pmod{4}$ for $n \in \{0, 1\}$.
We observe that $0$ is a solution, $1+25 = 26$ and $1 + 4 \cdot 25 = 101$ are solutions, $9 + 2 \cdot 25 = 59$ is a solution, $16 + 3 \cdot 25 = 91$ is a solution, and $4 + 15 \cdot 25 = 379$ is a solution.
Since $9$ is a solution, the smallest two-digit number whose cube ends with the original two-digit number is $\boxed{59}$.