Directional derivatives — implied continuity or not? 
Consider the function :
  $$f: \mathbb{R}^2 \rightarrow \mathbb{R} , (x,y) \mapsto
  \begin{cases}
        0 & \text{for } (x,y)=(0,0) \\
   \frac{x^2y^2}{x^2y^2+(y-x)^3}       & \text{for } (x,y) \neq (0,0)
  \end{cases} $$
  Is it differentiable and continuous at $0$?

I calculated the directional derivative which is equal to $0$ for all $u$ not equal to $0$. If all directional derivatives equal to $0$, does this imply differentiability and continuity? I cannot seem to find an answer online. But, if not, then how would i find if it is differentiable or not?
 A: The answer is no. Even if a function has a directional derivative for any direction, the possibility that the function is not continuous is still opened. The idea is that the directional derivative only captures the behavior of a function at a point along a line, so it could fail to catch its continuity or differentiability along other curves.
For example, consider
$$f(x,y)=\begin{cases} \frac{x^2y}{x^4+y^2} & \text{if }(x,y)\neq(0,0), \\ 0 & \text{if }(x,y)=(0,0).\end{cases}$$
and take $(x,y)=(t,t^2)$. Then $f(t,t^2)=\frac{1}{2}$. However, $f$ has a directional derivative at the origin for any direction.

In the case of your function, take $(x,y)=(t,t^2+t)$. Then
$$f(t,t^2+t) = \frac{t^4(t+1)^2}{t^4(t+1)^2+t^6}=\frac{1}{1+t^2(1+t)^{-1}}.$$
It tends to $1$ if $t\to 0$. However, $f(0,0)=0$, so $f$ is not even continuous at the origin.
A: Claim: The directional derivative does not exist with respect to the vector $u=(h,h)$ for $h\neq0$.
Proof:
$\begin{split}f'(0,u)&=\lim\limits_{t\to0}\frac{f(0+tu)-f(0)}{t}\\
&=\lim\limits_{t\to0}\frac{\frac{t^4h^4}{t^4h^4+t^3(h-h)^3}-0}{t}\\
&=\lim\limits_{t\to0}\frac{1}{t}\\
\end{split}$
This limit does not exist. Hense $f$ is not differentiable at(0,0).
