Applying implicit function theorem Suppose I have $2$ functions $\phi$ and $\psi$ and a $C^1$ function $F$ such that $F(\phi , \psi) = 0$
I know that we can write $\phi$ as a function of $\psi$ around a point if $F_\phi \neq 0 $ at that point. (Please mention some conditions I missed, if any.)
I am not sure I am applying the theorem correctly, so it would be helpful if someone can give a comprehensive example about this with some particular $\phi$ and $\psi$.
 A: If $\partial_1 F(x_0,y_0)\ne 0$ (also written as $\frac{\partial F}{\partial x}(x_0,y_0)\ne 0$), then exists $f$ in a nhood of $y_0$ s.t.
$$F(f(y),y) = 0$$
and we say that $x = f(y)$. Introducing $\phi$, $\psi$, this means that if
$$F(\phi(t),\psi(t)) = 0,$$then
$$\phi(t) = f(\psi(t)).$$
A: Simple example
Let $f: \mathbb R^3 \to \mathbb R$ defined by $f(x, y, z) = x + y + z$
Then the jacobian is $\begin{pmatrix} \frac{\partial f}{\partial x} & \frac{\partial f}{\partial y} & \frac{\partial f}{\partial z}\end{pmatrix} = \begin{pmatrix} 1 & 1 & 1\end{pmatrix}$
Let's look at the point $(1, 2, 3)$ specifically, notice that $f(1, 2, 3) = 6$. 
The sub matrix $\begin{pmatrix} \frac{\partial f}{\partial x}\end{pmatrix} = \begin{pmatrix} 1\end{pmatrix}$ is invertible at $(1, 2, 3)$. 
This means that there is an open neighborhood $V \subset \mathbb R$ that contains $1$ and a function $g: V \to \mathbb R^2$ such that $f(x, g(x)) = f(1, 2, 3) = 6$ for all $x \in V$, and in particular, $g(1) = (2, 3)$
Does this help or answer your question on how IFT works? Notice that in this case just like your case, $x,y, z$ could be functions of some other variable $t$. It does not matter, we derive $f$ with respect to $x, y, z$, as you should derive $F$ with respect to $\phi , \psi$. It does not matter that they themselves are functions.
A: The maps $\phi, \psi$ play no important role here. 
In fact you are looking at a (smooth) plane curve $F(x,y)=0$, an consider a function $\Phi= (\phi, \psi)$ with value in this curve. 
Locally this curve is the graph of a function $y= f(x)$ or $x=g(y)$, which yields the relation $\psi = f\circ \phi$ or $\phi = g\circ \psi$. 
Typically $F(x,y)= x^2+y^2-1$ and the curve is a circle ; in the domain $x>0$, $x= \sqrt {1-y^2}$, $\phi = \sqrt {1- \psi ^2}$, 
