Given the following scalar field $$f(x,y) = \begin{cases} \frac{y^3}{x^2+y^2} & (x,y)\ne(0,0) \\ 0 & (x,y)=(0,0) \end{cases}$$ find its directional derivative in the direction of $(3,2)$ at the point $(0,0)$.
First way I wanted to do this was with the gradient. However, as neither partial derivative exists at $(0,0)$, I need to use their limit form instead:
$$\frac{\partial}{\partial x}f(x,y)=\lim_{h\to 0}\frac{f(h,0)-f(0)}{h}=\lim_{h\to 0}\frac{f(h,0)}{h}=\lim_{h\to 0}\frac{0}{h^3}=0$$
$$\frac{\partial}{\partial y}f(x,y)=\lim_{k\to 0}\frac{f(0,k)-f(0)}{k}=\lim_{k\to 0}\frac{f(0,k)}{k}=\lim_{k\to 0}\frac{k^3}{k^3}=1$$
Yielding my gradient at $(0,0)$:
$$\nabla f = \vec{(0, 1)}$$
Using this to calculate the directional derivative with $u = (3,2)$, I get
$$\nabla_{u} f(0,0)=\nabla f_{(0,0)}\frac{u}{|u|}=\frac{1}{\sqrt{13}}(3,2)\cdot{}(0,1)=\frac{2}{\sqrt{13}}$$
However, if I use the directional derivative definition here:
$$\lim_{h\to 0}\frac{f(\vec{a}+h\vec{u})-f(\vec{a})}{h|u|}=\lim_{h\to 0}\frac{f(h(3,2))}{h\sqrt{13}}=\lim_{h\to 0}\frac{\frac{(2h)^3}{(3h)^2+(2h)^2}}{h\sqrt{13}}=\frac{8}{13\sqrt{13}}$$
which was the gradient answer but cubed. What should I expect from these answers? And what have I done wrong that yields these different but similar answers?
The function $f$ is NOT differentiable at $(0,0)$. According to the definition, the following limit should be zero, $$\lim_{(x,y)\to (0,0)}\frac{f(x,y)-f(0,0)-\nabla f_{(0,0)}\cdot (x,y)}{\sqrt{x^2+y^2}}=\lim_{(x,y)\to (0,0)}\frac{\frac{y^3}{x^2+y^2} -y}{\sqrt{x^2+y^2}}=\lim_{(x,y)\to (0,0)}\frac{-yx^2}{(x^2+y^2)^{3/2}}.$$ It turns out that this limit does not exist. It is $0$ along $y=0$ or $x=0$. On the other hand, for $y=x$, as $x\to 0^+$, we find $$\frac{-yx^2}{(x^2+y^2)^{3/2}}=\frac{-y^3}{(2y^2)^{3/2}}\to -\frac{1}{2^{3/2}}\not=0.$$ Therefore, without the differentiability condition, the equality $$\nabla_{u} f(0,0)=\nabla f_{(0,0)}\cdot \frac{u}{|u|}$$ may not be satisfied (as shown in your work).