For which $u$ does the derivative $f'(u,0)$ of $f(x,y)=xy/(x^2+y^2)$ exist? Let $f: \mathbb{R}^2\to \mathbb{R}$ be defined by setting
$$f(x,y)=\begin{cases}\frac{xy}{x^2+y^2}&(x,y)\neq (0,0)\\0 & (x,y)=(0,0)\end{cases}$$
(a) For which vectors $u\ne 0$ does $f'(0; u)$ exist? Evaluate it when it
exists.
(b) Do $D_1f$ and $D_2f$ exist at $0$?
(c) Is $f$ differentiable at $0$?
(d) Is $f$ continuous at $0$?
I have thought a lot about this problem: for (a), let $u\neq 0$ and $u=(h,k)$ then $\lim_{t\to 0}\frac{f(o+tu)-f(0)}{t}=\lim_{t\to 0}\frac{f(0+t(h,k))-f(0)}{t}=\lim_{t\to 0}\frac{hk}{t(h^2+k^2)}$ and so this limit exists if $h=0$ or $k=0$.
 A: b) Let $h= \in \Bbb{R}$ then $f(h,0)=f(0,h)=0$ so from this you can deduce that the partial derivatives at $(0,0)$ are zero.
d) Take the paths $(t,t)$ as $t \to 0$ and $(t,0)$ as $t \to 0$ and from these paths you will find different limits when approaching $(0,0)$, so you functions is not continuous at $(0,0)$ thus not differentiable at $(0,0)$
$\text{EDIT}$
Paths are curves, where in this case they pass through the point $(0,0)$.
If a limit exists then it is unique,thus uniqueness is a necessary condition for the existence of a limit at a given point.
If the limit of a function at a point $(x_0,y_0)$ exists then it will have the same value $l$ for every path we choose to approach the point $(x_0,y_0)$
So in this case the curves $g(t)=(t,t) \to (0,0)$ as $t \to 0$ and $h(t)=(t,0) \to (0,0)$ as $t \to 0$.
You can see this as approaching the origin from $\text{x-axis}$ namely where $y=0$ and $x$ varies,or approaching the origin from the diagonal,namely the line $y=x$
But these two approaches give two different results for the value of  $l$ thus the limit does not exist.
