Using the definition: $\lim_{x\to a}$ ${\left\Vert f(x)-f(a)-T(x-a) \right\Vert \over \left\Vert x-a \right\Vert}$ where ${T(x-a)}$ is the matrix of partial derivatives ${Df_a}$, I'd like to show that the function
$$f(\vec{x}) = {x_1x_2^2\over x_1^2 + x_2^2} \text{ if } \vec{x} \neq \vec 0,\quad f(\vec{0}) = \vec{0}$$
is not differentiable at $\vec{x} = \vec{0}$
The only thing is, I'm confused as to what the matrix of partial derivatives should be in this case since I've just been given this function.
Or is it just enough to show that two different paths approaching the point a, produce different values (but at the same time I think a differentiable function can still have discontinuous partial derivatives so that doesn't really help)
If it were, then the Jacobian is the zero matrix, and it amounts to see how the existence of the following limit leads to a contradiction: \begin{align*} \lim_{(h,k)\rightarrow(0,0)}\dfrac{\dfrac{hk^{2}}{h^{2}+k^{2}}}{\sqrt{h^{2}+k^{2}}}=0. \end{align*} Let $k=h$, $h>0$, then \begin{align*} \lim_{h\rightarrow 0^{+}}\dfrac{1}{\sqrt{2}h}\cdot\dfrac{h^{3}}{2h^{2}}=\dfrac{1}{2\sqrt{2}}\ne 0, \end{align*} here a contradiction.