calculuscontinuityderivativeslimits
In the following link, the function below is provided as an example of a function being continuous and all directional derivatives existing. Yet, it is not differentiable. How do I prove that this is not differentiable?
$$
f(x,y)=
\begin{cases}
\frac{y^3}{x^2+y^2},& \text{if } (x,y) \neq (0,0)\\
0, & \text{otherwise}
\end{cases}
$$
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 $(1,1)$.
There are no directional derivatives in nearly all directions. Consider, in particular, along the line $y=x$. $f(x,y)$ is a constant times the absolute value function.
When a function of two variables is differentiable, then there is a tangent plane to the surface $z=f(x,y)$, and there are directional derivatives in all directions. This one doesn't have directional derivatives except in two directions, and there's no tangent plane to the surface $z=f(x,y)$.
Best Answer
The definition of the directional derivative at $0$ is $$df(v)=\lim\limits_{t\to 0}\frac{f(tv)-f(0)}{t}$$ if $v=(a,b)$ then
$$\frac{f(tv)-f(0)}{t}=\frac{\frac{t^3a^3}{t^2a^2+t^2b^2}}{t}=\frac{a^3}{a^2+b^2}$$
So the directional derivative is $$df(a,b)=\frac{a^3}{a^2+b^2}$$ Now the condition for differentiablity is that $df$ is a linear function. This is not the case so the function is not differentiable, although all directional derivatives exist.
I had a minor realization about this recently, if you take for example $z=f(x,y)$ and graph it as a surface and if directional derivatives of $f$ exist then then the surface will have slopes in all directions. So there are well defined tangent lines in all directions. However the function is differentiable only if all those tangent lines lie on the same plane. If you graph this function in wolfram alpha you can see that this is not the case, as was also shown above.