Let $S^{1}$ be the unit circle in $\mathbb{R}^{2}$, which we can see as $f^{-1}(0)$, where $$ f(x,y) = x^{2} + y^{2} -1 $$ The differential of this function at any point is $$ Df(x,y) = \begin{pmatrix} 2x & 2y \\ \end{pmatrix} $$ The Implicit Function Theorem tells us that for $(x,y) \in S^{n-1}$, if $2y \neq 0$, then we have a function $g$ that maps a neighborhood around $x$ to a neighborhood around $y$, and $(x, g(x)) \in S^{n-1}$ for $x$ in this neighborhood, and it is unique in this neighborhood.
Now, it is clear that getting a unique function like this is not possible at $(1,0)$, since we can see that we will need the functions $y = \sqrt{1-x^{2}}$ and $y = - \sqrt{1-x^{2}}$ to define all $y$ in any neighborhood of $(1,0)$ as a function of $x$. So the Implicit Function Theorem doesn't hold, which doesn't contradict anything, since $2y = 0$ here so our assumption fails.
Now, my question is on the intuition of this. What is the connection between the fact that $2y = 0$ at $(1,0)$ and that we can't get. a unique function based on the observation I have given above? How does $2y$ not being $0$ tell me that the function won't be unique in any neighborhood of my point?
The Implicit Function Theorem requires the function to have a nonzero partial derivative in the $y$ direction. Loosely speaking, it ensures that if you are on the curve and slightly nudge $x$ to the left or right, there is always another slight nudge of $y$ up or down such that you are still on the curve.
Now consider the point $(1,0)$. If we nudge $x$ ever so slightly to the right, let's say to the point $1 + h$, with $h \to 0$. Can we find a value of $y$ such that we are still on the curve? That means that we would need $(1+h)^2 +y^2-1 =0$ to hold, which (if we expand the brackets) means that $1+2h+h^2+y^2-1=0$, or $y^2=-2h-h^2 = -h(h+2)$. However, since $h$ is positive, $y^2$ must be negative, which is not possible in the real numbers. Therefore we cannot find a $y$ such that we are still on the curve if w slightly nudge $x$ to the right.
The reason why this is the case, is because the curve that satisfies $f(x,y)=0$ at the point $(1,0)$ is vertical. Another way of saying the curve is vertical, is the statement that the partial derivative $\frac{\partial f}{\partial y}(1,0)=0$, which why the Implicit Function Theorem requires that $\frac{\partial f}{\partial y}(x,y)\neq0$ at the point of interest.