Show that $\frac{x^3}{x^2+y^2}$ is not differentiable at $(0,0)$, even though all directional derivatives exist

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 $(1,1)$.

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.