Why does existence of directional derivatives not imply differentiability? In my notes I have:
$$Df\big|_{\mathbf{a}}(\mathbf{h})=\lim_{t\to 0}\frac{f(\mathbf{a}+t\mathbf{h})-f(\mathbf{a})}{t}$$
It says that even if this limit exists for all $\mathbf{h}$, we do not necessarily have differentiability. Is this because the $Df\big|_{\mathbf{a}}$ we get from this is not necessarily linear, or even continuous, in $\mathbf{h}?$ Can I say that if the result map is linear & continuous then we have differentiability?

I ask because I am doing a problem which asks to check differentiability of some functions $f:\mathbb{R}^2\to\mathbb{R}$. One of them is:
$$f(x,y)=xy\left(\frac{x^4-y^4}{x^4+y^4}\right)\qquad f(0,0)=0$$
which I thought was differentiable at $\mathbf{0}$ since $Df\big|_{\mathbf{0}}(\mathbf{h})=0$ for all $\mathbf{h}$, but I am not sure this is enough.
 A: Existence of every directional derivative of $f$ at a point $p$ does not even imply $f$ is continuous at $p$. (!) For example, let $E$ denote the parabola $y = x^{2}$ with the origin removed, and let $f$ be the characteristic function of $E$, namely
$$
f(x, y)
  = \begin{cases}
    1 & y = x^{2},\ x \neq 0, \\
    0 & \text{otherwise.}
  \end{cases}
$$

Understanding this example thoroughly should clarify your doubts about your assigned question. The point is, for every line $\ell$ through the origin, $\ell$ contains an open interval about $(0, 0)$ that misses $E$.[*] Consequently, $f$ is locally constant along every line through the origin, so its directional derivative in an arbitrary direction vanishes.
Conceptually, controlling the behavior of a function along lines through a point need not give control over non-lines through that point. That's why limits at $p$ are defined by looking at the behavior of $f$ in open disks about $p$.
[*] Precisely, for every line $\ell$ through $(0, 0)$, there exists an $r > 0$ such that $\ell$ does not meet $E$ inside the open disk of radius $r$ centered at the origin.
A: It is true that the existence of that limit does not imply differentiability. Existence and continuity of all the partial derivatives of $f$ in $a$ implies differentiability of $f$ in $a$. In your notation, we get the partial derivatives by calculating the limit $Df|_{a}(\textbf{e}_i)$, where $\{ \textbf{e}_i \}$ denotes the standard basis of $\mathbb{R}^n$. However, I do not recommend this notation, because $Df|_{a}$ often denotes the total derivative in $a$, whose existence you need to prove! $\textbf{If}$ the total derivative $Df|_{a}$ exists, then the directional derivative in the direction of $\textbf{h}$ is given by $Df|_{a}(\textbf{h})$, the total derivative in $a$ evaluated at $\textbf{h}$.
So, to solve your problem, I would recommend calculating the partial derivatives of $f$, and checking their continuity in $\textbf{0}$. For this you will need to calculate the partial derivatives of $f$ in $\textbf{0}$ as well as in a punctured neighbourhood of $\textbf{0}$, that is, in some $B(\textbf{0}, \epsilon)-\{\textbf{0}\}$.
A: Hint:
A function $f:U\subset\mathbb{R}^m\rightarrow \mathbb{R}^n$ is differentiable at $a \in U$ if there exists a linear transformation $T_a:\mathbb{R}^m\rightarrow \mathbb{R}^n$ such that
$$f(a+h) = f(a) + T_a\cdot h + r(h), \quad\text{ where } \lim_{h\rightarrow0} \frac{r(h)}{\|h\|} = 0.$$
So, yes, $f'(a) \equiv T_a$ must be a linear transformation. Try to find a point $x \in U$ such that $f'(x)$ is not a linear transformation, i.e. $f'(x)\cdot (\alpha h + k) \neq \alpha f'(x)\cdot h + f'(x)\cdot k$.
