Show that $\frac{x^3}{x^2+y^2}$ is not differentiable at $(0,0)$, even though all directional derivatives exist Consider the function :
$$f: \mathbb{R}^2 \rightarrow \mathbb{R} , (x,y) \mapsto
  \begin{cases}
        0 & \text{for } (x,y)=(0,0) \\
   \frac{x^3}{x^2+y^2}       & \text{for } (x,y) \neq (0,0)
  \end{cases} $$
Show that $f$ not differentiable at $(0,0)$ but all directional derivatives exist.
I don't know how to tackle this problem. Can someone give some hints or solve the problem? Thanks :)
 A: Since someone has already shown that all directional derivatives exist, I will only argue why $f$ is not differentiable at $0$.
The Jacobi Matrix $A:=Df(0,0)=(1,0)$. Therefore if $f$ is differentiable $$\lim_{|\epsilon| \to 0}\frac{f(0+\epsilon)-f(0)-A\epsilon}{|\epsilon|}=0 .$$
Since $f(0)=0$ and $A=(1,0)$ this is equivalent to, 
$$\lim_{|\epsilon| \to 0}\frac{f(\epsilon)-(1,0)\epsilon}{|\epsilon|}=0 $$
Let $(x_k)_{k \in \mathbb{N}} \subset \mathbb{R}^2$ be a series with $x_k=(a_k,b_k)$ and $|x_k| \to 0$ for $  (k \to \infty)$. Therefore if $f$ is differentiable.
$$0=\lim_{k \to \infty}\frac{f(x_k)-(1,0)x_k}{|x_k|}=\lim_{k \to \infty}\frac{\frac{a_k^3}{a_k^2+b_k^2}-a_k}{\sqrt{a_k^2+b_k^2}}= \lim_{k \to \infty}\frac{-a_k b_k^2}{\sqrt{a_k^2+b_k^2}^3}$$
If we set $x_k=(a_k,b_k)= (\frac {1}{k\sqrt{3}},\frac{\sqrt{2}}{k\sqrt{3}})$ then $|x_k|=\sqrt{a_k^2+b_k^2}=1/k$ and
$$\lim_{k \to \infty}\frac{-a_k b_k^2}{\sqrt{a_k^2+b_k^2}^3}=\lim_{k \to \infty}\frac{-\frac {1}{k\sqrt{3}} \frac{2}{3k^2}}{(1/k)^3}=\lim_{k \to \infty}-k^3\frac{2}{k^3\sqrt{3}^3}=\frac{-2}{\sqrt{3}^3}\neq 0$$
We get a contradiction therefore $f$ is not differentiable in 0. 
A: The directional derivative is $$\lim_{t\to0} \frac{f((0,0)-(th_1,th_2)) - f(0,0)}{t}= \lim_{t\to 0} \frac{(th_1)^3}{t(t^2 h_1^2 + t^2 h_2^2)} = \frac{h_1^3}{h_1^2+h_2^2}$$
If we set $h=(1,0)$ and $h = (0,1)$, we get $\partial f/\partial x$ and $\partial f/\partial y$ respectively. So $Df(0,0) = (1,0).$
$$\lim_{h \to 0} \frac{f(0 + h) - f(0) - Df(0,0)h}{|h|} = \lim_{h \to 0} \frac{ \frac{h_1^3}{h_1^2 + h_2^2} - h_1 }{\sqrt{h_1^2 + h_2^2}} = \lim_{h \to 0} = \frac{-h_2^2h_1}{(h_1^2 + h_2^2)^{3/2}}.$$
If you let $h_1 = h_2m$ where $m \in \mathbb{R}$, then you can show this limit doesn't exist. Therefore $f$ is not differentiable. 
A: Take some (unitary) $u$. Then show that $$f'(\vec 0;u)=\lim_{h\to 0 }\frac{f(h\cdot u)-f(\vec 0)}h$$ always exists for any choice of $u$. You'll be dealing with $$\mathop {\lim }\limits_{h \to 0} \frac{\frac{{{h^3}u_1^3}}{{{h^2}(u_1^2 + u_2^2)}}}{h}$$
where $u=(u_1,u_2)$.
If your function were differentiable at the origin, then we would have $$f'(\vec 0)(u)=f'(\vec 0;u)$$ where the right hand side is the directional derivative at $\vec 0$ with direction $u$, and the left hand side is the total derivative at $(0,0)$ at evaluated at $u$. Now, $f'(\vec 0)$ would be linear, so $$f'(\vec 0)(1,1)=f'(\vec 0)(0,1)+f'(\vec 0)(1,0)$$
The right hand side is just the partial derivatives at the origin, which gives $0+1=1$. What does the left hand side give? Remember it is just the directional derivative at $\vec 0$ with direction $(1,1)$.
A: When $x>0$, then $$df\ \frac{(1,k)}{\sqrt{1+k^2}} = \lim_x\ \frac{ f(x,kx) -0 }{ |(x,kx)|}
 = \frac{1}{\sqrt{1+k^2}^3} $$
Hence we consider directional derivatives at $(0,0)$ : $$
 df\ (1,0)=1\geq df\ \frac{(1,k)}{\sqrt{1+k^2}} \geq df\ (0,1)=0$$
If $f$ is differentiable at $(0,0)$, then directional derivative
 is given by the formula $df\ v ={\rm grad}\
 f\cdot v$.
Hence ${\rm grad}\ f = (1,0)$ so that $$
 (1,0) \cdot
 \frac{(1,k)}{\sqrt{1+k^2} } \neq df\ \frac{(1,k)}{\sqrt{1+k^2}} =
 \frac{1}{\sqrt{1+k^2}^3} $$
