What could be conditions on $k\in\mathbb{C}[x,y,z]$ that would ensure that any polynomial $f\in\mathbb{C}[x,y,z]$ that is algebraically dependent of $k$ is indeed a polynomial in $k$, ie $f\in\mathbb{C}[k]$. References?
$\def\CC{\mathbb{C}}$A necessary and sufficient condition is that $k$ cannot be written as $h(\ell(x,y,z))$ for $h \in \CC[t]$ of degree $>1$ and $\ell \in \CC[x,y,z]$. Clearly, this is a necessary condition since, if $k = h(\ell)$, then $k$ and $\ell$ are integrally dependent. We now prove sufficiency. Let $A = \CC[k]$ and let $B$ be the integral closure of $A$ in $\CC[x,y,z]$. I claim that $B \cong \mathbb{C}[\ell]$ for some $\ell \in \mathbb{C}[x,y,z]$. First of all, $B$ is finite over $A$ and so finitely generated and onedimensional. Also, it is normal. So $\mathrm{Spec}(B)$ is a smooth curve of some genus and some number of punctures. Since $B \subset \CC(x,y,z)$, the curve is unirational which, for curves, is the same thing as rational. So $\mathrm{Spec}(B)$ is genus zero. Also, $B \subset \CC[x,y,z]$ so $B$ has no units other than $\CC^{\times}$ and we see that $\mathrm{Spec}(B)$ has only one puncture. In short, $\mathrm{Spec}(B) \cong \mathbb{A}^1$ and $B \cong \CC[\ell]$. In particular, we have $k=h(\ell)$ for some $h \in \CC[t]$. If $h$ has degree $\geq 2$, as noted above, than $k$ and $\ell$ are algebraically dependent and $\ell \not \in \CC[k]$. Conversely, suppose that $h$ has degree $1$, in which case $B = A$. I claim that, if $f$ and $k$ are algebraically dependent, then $f \in \CC[k]$. Proof: If $f$ and $k$ are algebraically dependent then $p(k) \cdot f$ is integral over $\CC[k]$ for some polynomial $p$. So $p(k) f \in B = \CC[k]$ and we deduce that $f \in \mathrm{Frac} \ \CC[k]$. If $f$ is in $\mathrm{Frac}\ \CC[k]$ but not $\CC[k]$, then the ring $\CC[k,f]$ will contain a unit not in $\mathbb{C}^{\times}$, contradicting that $k$ and $f$ are both in $\CC[x,y,z]$. 

