The condition is not sufficient. Consider $f(x,y)=|y|^\alpha e^{-\frac{x^2}{y^2}}$ for $(x,y)\neq (0,0)$ and $f(0,0)=0$ with $\alpha\in (0,1)$. The function is continuous at $(0,0)$ as $0\geq f(x,y)\neq |y|^\alpha $ for $(x,y)\neq (0,0)$.
The directional derivative at $(0,0)$ is specified by the versor $v(\theta)=(\cos\theta,\sin\theta)$. Then
$$\frac{f(t\cos\theta, t\sin\theta)}{t}=\frac{|t\sin\theta|^\alpha}{t} e^{-\operatorname{cotg}^2\theta}, $$
if $\theta\neq 0,\pi$ and
$$\frac{f(t,0)}{t}=0. $$
if $\theta = 0,\pi$. It follows that $f_x(0,0)=0$ identically.
However, if $\theta$ is different from $0$ and $\pi$, the limit $\lim_{t\rightarrow 0}\frac{f(t\cos\theta, t\sin\theta)}{t}$ does not converge as $\alpha-1<0$.
A final remark: if a given function is differentiable at a given point, then all directional derivatives- and so also the partial derivatives- at that point exist, however.
Directional derivative of $f$ at $(0,0)$ in the direction $(v,w)$ can be understood as follows:
- form the composition $g(t) = f(0+tv,0+tw)$
- study the existence of $g'(0)$.
In your case, $g(t) = |(2v+w)t| = |2v+w| |t|$. Is this differentiable at $0$?
- Yes when $2v+w=0$ (why?)
- No when $2v+w\ne 0$ (relate to $h(t)=|t|$, a canonical example of a nondifferentiable function)
Best Answer
The direction derivatives at the origin do not exist. If $v=(a,b,c)\neq(0,0,0)$, then the directional derivative at the origin of $f$ with respect to the direction given by $v$ is$$\lim_{t\to0}\frac{f(ta,tv,tc)-f(0,0,0)}t=\lim_{t\to0}\frac{\lvert t\rvert}t\sqrt{a^2+b^2+c^2}$$and, since $\sqrt{a^2+b^2+c^2}\neq0$, this limit doesn't exist.