Directional derivatives — implied continuity or not?

Question

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?

Answer 1

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.

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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.

Answer 2

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).