The Exercise 9 Sec. 13.3 Fitzpatrick's Calculus says:
Define the function $f : \mathbb{R}^2 \to \mathbb{R}$ by $$f(x,y) = \left\{ \begin{array}{ll} (x/|y|)\sqrt{x^2+y^2} & \mbox{if } y \ne 0 \\ 0 & \mbox{if } y = 0. \end{array} \right.$$ a. Prove that the function $f : \mathbb{R}^2 \to \mathbb{R}$ is not continuous at the point $(0,0)$.
b. Prove that the function $f : \mathbb{R}^2 \to \mathbb{R}$ has directional derivatives in all directions at the point $(0,0)$.
c. Prove that if $c$ is any number, then there is a vector $p$ of norm $1$ such that $$\frac{\partial f}{\partial p} (0,0) = c.$$
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I can't solve none of three parts of the question:
a. I try to find a counterexample that the given function is not continuous at the point $(0,0)$. But approaching with any sequence of pairs of numbers ${\{x_k,y_k}\}$ like ${\{1/k,1/k}\}$, etc it always leads to $(0,0)$, so how the function is not continuous when it seems to be?
b. The given function has directional derivatives in all directions at the point $(0,0)$ iff the limit $$\lim_{t \to 0} \dfrac{f((0,0)+tp)-f(0,0)}{t}$$ exists; then, as $f(0,0)=0$, the above limit equals $$\lim_{t \to 0} (t/|t|)(p_1/|p_2|)\sqrt{{p_1}^2+{p_2}^2} $$ which doesn't exist because of the factor $t/|t|$, so how the function has directional derivatives in all directions at the point $(0,0)$?
c. If b. is solved so is this part.
No continuity at $(0,0)$
You have for $t \neq 0$: $$\left\vert f(t,t^2) \right\vert = \frac{\sqrt{t^2 + t^4}}{\vert t \vert} \ge 1$$ and $\lim\limits_{t \to 0} (t,t^2) = (0,0)$. Therefore you can't have $\lim\limits_{t \to 0} f(t,t^2) = 0$, in contradiction with $f(0,0) = 0$, proving that $f$ is not continuous at $(0,0)$.
Directional derivatives at $(0,0)$.
I agree with your comments. The limit exists only for $\lim\limits_{t \to 0+}$.