I don't know very much about differential geometry, so this may be trivial but I will try to make this question as clear as I can.
Suppose I have a $d$ dimensional smooth manifold $\mathcal{M}^d$, with a smooth maximal atlas $\mathcal{A}$. In the neighborhood of a point $m$ I have a chart $(U,\phi) \in \mathcal{A}$ that gives some local coordinates for my manifold, say $\phi(m) = (x^1(m), ..., x^d(m))$ and $\phi : U \subseteq \mathcal{M}^d \rightarrow \mathbb{R}^d$ is a homeomorphism. Suppose also that I have a smooth function defined on my manifold, $f: \mathcal{M}\rightarrow \mathbb{R}$.
My question is the following: Are there any conditions that guarantee that the chart $(V, \psi)$, with $m \in U \cap V$, $\psi = (x^1, ..., f)$ is also in $\mathcal{A}$?
In other words, how can I be sure that I can replace one of my coordinates with the smooth function $f$?
Take a look at Tu's Introduction to manifolds, Section 6.3 on the Inverse Function Theorem.
The way the IFT is written there is that some local functions $(f_1,\dots,f_n):U\to\mathbb R^n$ define a coordinate chart around $p\in U$ iff there's a chart $(x_1,\dots,x_n)$ around $p$ such that $$ \det\begin{bmatrix}\dfrac{\partial f_i}{\partial x^j}\end{bmatrix}\neq 0 $$
So we get a sufficient a condition: $(x_1,\dots,x_{n-1},f)$ defines a chart if $$ \det \begin{bmatrix} \dfrac{\partial x_1}{\partial x^1} &\cdots &\dfrac{\partial x_1}{\partial x^{n-1}}&\dfrac{\partial x_1}{\partial x^n}\\ \vdots & \ddots &\vdots & \vdots \\ \dfrac{\partial x_{n-1}}{\partial x^1} &\cdots &\dfrac{\partial x_{n-1}}{\partial x^{n-1}}&\dfrac{\partial x_{n-1}}{\partial x^n}\\ \dfrac{\partial f}{\partial x^1}&\cdots&\dfrac{\partial f}{\partial x^{n-1}}&\dfrac{\partial f}{\partial x^n} \end{bmatrix} = \det \begin{bmatrix} 1 & \cdots &0&0\\ \vdots & \ddots &\vdots & \vdots \\ 0 &\cdots &1&0\\ \dfrac{\partial f}{\partial x^1}&\cdots&\cdots&\dfrac{\partial f}{\partial x^n} \end{bmatrix} = \dfrac{\partial f}{\partial x^n} \neq 0. $$