Theorem 1.9.1. Let $U \in \R^n$ be an open, let $f: U \to \R$ be differentiable, and let the segment $[a,b]$ joining $a$ to $b$ be contained in $U$. Then there exists $c_0 \in [a,b]$ such that
$$ f(b) -f(a) = Df(c_0). $$
It is possible for all partial derivatives of $f$ to exist, and even all directional derivatives, and yet $f$ to not be differentiable. In this case, the Jacobian exists but does not represent the derivative.
Example:
$$ f\begin{pmatrix}x\\y\end{pmatrix}\begin{cases} \frac{x^2y}{x^2+y^2} & \text{if }\begin{pmatrix} x\\y\end{pmatrix}\neq \begin{pmatrix}0\\0\end{pmatrix}\\ 0 &\text{if }\begin{pmatrix} x\\y\end{pmatrix}= \begin{pmatrix}0\\0\end{pmatrix} \end{cases} $$
This is indeed continuous at the origin. First we prove that all directional derivatives exist.
Let $\vec v = \begin{bmatrix}x\\y\end{bmatrix}$. Recall from Chapter 7 that the directional derivative is equal to
$$ \lim_{h\to 0}\frac{f(a+h\vec v)-f(a)}{h} $$
For $f$, we get
$$ \lim_{h\to 0}\frac{f\begin{pmatrix}hx\\hy\end{pmatrix} - f\begin{pmatrix}0\\0\end{pmatrix}}{h} $$
$$ =\lim_{h\to 0}\frac{1}{h}\frac{h^3x^2y}{h^2(x^2+y^2)} $$
$$ = \frac{x^2y}{x^2+y^2} $$
However, the total derivative does not exist. The derivatives in different directions are not equal. For example, take the derivative in the direction of vector $\begin{bmatrix}1\\1\end{bmatrix}$.
$$ \lim_{t\to 0}\frac{f\begin{pmatrix}\begin{pmatrix}0\\0\end{pmatrix}+t\begin{bmatrix}1\\1\end{bmatrix}\end{pmatrix}-f\begin{pmatrix}0\\0\end{pmatrix}}{t}= \lim_{t\to 0}\frac{t^3}{2t^3} = \frac{1}{2} $$
This is not equal to the Jacobian matrix times the vector $\begin{bmatrix}1\\1\end{bmatrix}$.