Being algebraically independent is indeed a necessary and sufficient condition for the image of $f$ to be dense.

As $f\colon\mathbb{C}^n\to\mathbb{C}^n$ is regular, its image is constructible and, in particular, contains a non-empty open subset of its closure (under the Zariski topology). See Theorem 10.2 of J.S. Milne's algebraic geometry notes. If the Zariski closure of $f(\mathbb{C}^n)$ is all of $\mathbb{C}^n$, then $f(\mathbb{C}^n)$ contains a Zariski open set and is dense in the standard topology. Otherwise, the Zariski closure of $f(\mathbb{C}^n)$ is the zero set of a nontrivial ideal $I\subset\mathbb{C}[X_1,\ldots,X_n]$ and you can take any $F\in I\setminus\{0\}$ to see that $f_i$ are algebraically dependent.

Being algebraically independent is indeed a necessary and sufficient condition for the image of $f$ to be dense.

As $f\colon\mathbb{C}^n\to\mathbb{C}^n$ is regular, its image is constructible and, in particular, contains a non-empty open subset of its closure (under the Zariski topology). See Theorem 10.2 of J.S. Milne's algebraic geometry notes. If the Zariski closure of $f(\mathbb{C}^n)$ is all of $\mathbb{C}^n$, then $f(\mathbb{C}^n)$ contains a Zariski open set and is dense in the standard topology. Otherwise, the Zariski closure of $f(\mathbb{C}^n)$ is the zero set of a nontrivial ideal $I\subset\mathbb{C}[X_1,\ldots,X_n]$ and you can take any $F\in I\setminus\{0\}$ to see that $f_i$ are algebraically dependent. In fact, this argument works for regular maps $f\colon V\to\mathbb{C}^n$ for any variety $V$. There is no need to restrict the domain to be $\mathbb{C}^n$.

Being algebraically independent is indeed a necessary and sufficient condition for the image of $f$ to be dense. By the standard embedding into projective space $\mathbb{C}^n\to\mathbb{P}^n$, we can extend $f$ to a regular map $g\colon\mathbb{P}^n\to\mathbb{P}^n$. As projective space is a complete variety, the image of $g$ is Zariski-closed. If the image is not all of $\mathbb{P}^n$ then the image of $f$ is not Zariski dense, so lies in the zero set of some nontrivial ideal $I\in\mathbb{C}[X_1,\ldots,X_n]$. You can take any $F\in I\setminus\{0\}$ to show that the $f_i$ are not algebraically independent. On the other hand, if the image of $g$ is the whole space, then the image of $f$ is all of $\mathbb{C}^n$ minus $g(\mathbb{P}^n\setminus\mathbb{C}^n)$, which is a dimension of variety less than $n$. In that case, the image of $f$ is dense.

Being algebraically independent is indeed a necessary and sufficient condition for the image of $f$ to be dense.

As $f\colon\mathbb{C}^n\to\mathbb{C}^n$ is regular, its image is constructible and, in particular, contains a non-empty open subset of its closure (under the Zariski topology). See Theorem 10.2 of J.S. Milne's algebraic geometry notes. If the Zariski closure of $f(\mathbb{C}^n)$ is all of $\mathbb{C}^n$, then $f(\mathbb{C}^n)$ contains a Zariski open set and is dense in the standard topology. Otherwise, the Zariski closure of $f(\mathbb{C}^n)$ is the zero set of a nontrivial ideal $I\subset\mathbb{C}[X_1,\ldots,X_n]$ and you can take any $F\in I\setminus\{0\}$ to see that $f_i$ are algebraically dependent.