Let's determine those fractions where both the numerator and the denominator are two-digit numbers, the second digit of the numerator is the same as the first digit of the denominator, and the value of the fraction does not change if we omit the two identical digits.
Solution. The fraction can be written as:

$$
\frac{\overline{x y}}{\overline{y z}}=\frac{10 x+y}{10 y+z}=\frac{x}{z}
$$

Since the number is two-digit, $x$ and $y$ cannot be 0, and $z$ cannot be 0 either, as the denominator of a fraction cannot be 0. Thus, $1 \leq x, y, z \leq 9$. Let's transform the equation:

$$
\begin{aligned}
(10 x+y) \cdot z & =(10 y+z) \cdot x \\
10 x z+y z & =10 x y+x z \\
9 x z & =y \cdot(10 x-z)
\end{aligned}
$$

The left side is divisible by 9, so the right side must also be divisible by 9.

There are three possibilities.

(1) $9 \mid y$. This can only happen if $y=9$. In this case:

$$
\begin{aligned}
x z & =10 x-z \\
z(x+1) & =10 x \\
z & =\frac{10 x}{x+1}
\end{aligned}
$$

Substituting the possible values of $x$, $z$ is an integer only when $x=1,4$ and 9 (the values of $z$ are 5, 8, and 9, respectively). These solutions are all valid, so the corresponding fractions are: $\frac{19}{95}, \frac{49}{98}, \frac{99}{99}$.

(2) $9 \mid(10 x-z)=9 x+(x-z)$, which means $9 \mid x-z$. Since $0 \leq|x-z| \leq 8$, the only solution is when $x=z$, and all these solutions are valid. Since the value of the right-side fraction is always 1 in this case, only $x=y=z$ is possible. Therefore, the corresponding fractions are: $\frac{11}{11}, \frac{22}{22}, \frac{33}{33}, \frac{44}{44}, \frac{55}{55}, \frac{66}{66}, \frac{77}{77}, \frac{88}{88}, \frac{99}{99}$.

(3) $3 \mid y$ and $3 \mid(10 x-z)$. Here, there are two possibilities: $y=3$ and $y=6$ (we have already considered $y=9$).

If $y=3$, then

$$
\begin{aligned}
9 x z & =3(10 x-z), \\
3 x z & =10 x-z, \\
z(3 x+1) & =10 x, \\
z & =\frac{10 x}{3 x+1} .
\end{aligned}
$$

Here, $z$ is an integer only when $x=3$, in which case $x=y=z=3$, giving the fraction $\frac{33}{33}$ as a solution.

If $y=6$, then

$$
\begin{aligned}
9 x z & =6(10 x-z) \\
3 x z & =20 x-2 z \\
z(3 x+2) & =20 x \\
z & =\frac{20 x}{3 x+2} .
\end{aligned}
$$

Here, the values $x=1,2$ and 6 give integer values for $z$, which are 4, 5, and 6, respectively. The corresponding fractions are: $\frac{16}{64}, \frac{26}{65}, \frac{66}{66}$. Therefore, these are the solutions:

$$
\frac{19}{95}, \frac{49}{98}, \frac{11}{11}, \frac{22}{22}, \frac{33}{33}, \frac{44}{44}, \frac{55}{55}, \frac{66}{66}, \frac{77}{77}, \frac{88}{88}, \frac{99}{99}, \frac{16}{64}, \frac{26}{65}
$$