$I(X;f(Y, Z))$ can depend on the $Z$ (or more specifically the distribution of $Z$). Consider the following example, $Z \sim B(p)$, $X \sim B(0.5)$, $Y=X$ satisfying $X,Y \perp \!\!\! \perp Z$. Let $F=max(Y, Z)$ a variable from the output of a deterministic function of $Y, Z$. $X, F$ have the following joint distribution,

1. $P(X=0,F=0) = (1-p)/2$.
2. $P(X=0,F=1) = p/2$.
3. $P(X=1,F=0) = 0$.
4. $P(X=1,F=1) = 1/2$.

$I(X;F) = \log(2) + \frac{p}{2}\log(p) - \frac{1+p}{2}\log(1+p)$ which is dependent of $p$.

$I(X;f(Y, Z))$ can depend on the $Z$ (or more specifically the distribution of $Z$). Consider the following example, $Z \sim B(p)$, $X \sim B(0.5)$, $Y=X$ satisfying $X,Y \perp \!\!\! \perp Z$. Let $F=\max(Y, Z)$ be a variable from the output of a deterministic function of $Y, Z$. $X, F$ have the following joint distribution,

1. $P(X=0,F=0) = (1-p)/2$.
2. $P(X=0,F=1) = p/2$.
3. $P(X=1,F=0) = 0$.
4. $P(X=1,F=1) = 1/2$.

$I(X;F) = \log(2) + \frac{p}{2}\log(p) - \frac{1+p}{2}\log(1+p)$ which is dependent on $p$.