If $\{x,y\}$ and $\{y,z\}$ are both algebraically dependent, then is also $\{x,z\}$ algebraically dependent?
The important thing is: if they are, then algebraic dependence is an equivalence relation (because reflexivity and simmetry are trivially true).
If yes, do their equivalence classes have some special meaning or parametrization?
If we are referring to elements of a common field $F\supseteq k(x,y,z)$ and algebraic dependence is over some fixed subfield $k$, then yes, because \begin{align}[k(x,z):k(x)]\le& [k(x,y,z):k(x)]=[k(x,y,z):k(x,y)][k(x,y):k(x)]\le\\\le& [k(z,y):k(y)][k(x,y):k(x)]\end{align}
Where $[k(z,y):k(y)]\ge [k(x,y,z):k(x,y)]$ because if $f\in k(y)[T]$ is a primitive polynomial for $z$ over $k(y)$, then $f\in k(x,y)[T]$ in the obvious way and it's a fortiori a multiple of some primitive polynomial for $z$ over $k(x,y)$.
When $x$ isn't algebraically dependent with $0$ over $k$, the equivalence class of the elements of $F$ that are algebraically dependent with $x$ is the subfield of the elements of $F$ that are algebraic over $k(x)$, minus the subfield of the elements of $F$ that are algebraic over $k$. The class of $0$ is the subfield of the elements of $F$ that are algebraic over $k$.