Let $f(x,y)= \begin{cases} \frac{2x^2y}{x^2+y^2} & \text{if } (x,y) \neq (0,0) \\ 0 & \text{if } (x,y)=(0,0) . \end{cases}$
(i) Prove the directional derivatives of $f$ exist in any direction at the point $(0,0)$.
(ii)¿Is $f$ continuous on $(0,0)$?
For (i) I took $u=(u_{1},u_{2}) \in \mathbb{R^{2}}$ such $\|u\|=1$ and $0=(0,0)$. So $$\lim_{t \to 0} \frac{f(0+ t u)-f(0)}{t}=\lim_{t \to 0}\frac{2(tu_{1})^2 (tu_{2})}{(tu_{1})^4+(tu_{2})^2}=\lim_{t \to 0}\frac{(t^2 u_{1}^2)(t^2 u_{1}^{2}) tu_{2}}{t^{4}u_{1}^{4}+t^{2} u_{2}^{2}}$$. But I cannot find the limit when $t$ approaches to $0$. Which by finding them I would prove the directional derivative exist at every direction right?
For (ii) I got the intuition $f$ is not continuous at $(0,0)$ so I took $\lbrace (\frac{1}{k},\frac{1}{k}) \rbrace_{k \in \mathbb{N}}$ and $\lbrace (\frac{1}{k},0) \rbrace_{k \in \mathbb{N}}$ which are two different sequences in $\mathbb{R}^{2}$ converging to $(0,0)$. However, $\lbrace f(\frac{1}{k},\frac{1}{k}) \rbrace_{k \in \mathbb{N}}=\lbrace \frac{1}{k^{2}} \rbrace_{n \in \mathbb{N}} \to 0$ and $\lbrace f(\frac{1}{k},0) \rbrace_{k \in \mathbb{N}} \to 0$. So maybe my intuition was not right?
Also I've have approach $(0,0)$ through $(x,mx)$ and when this and gives me 0 as this value approaches to zero bit I cannot find another trajectory approaching through $(0,0)$ which gives me another value distinct to $0$. :(
Can anyone help me end the proof of continuity or not continuity, please?
Using polar coordinates, $\mid\frac{2x^2y}{x^2+y^2}\mid=\mid\frac{2r^2\cos^2\theta r\sin\theta}{r^2\cos^2\theta +r^2\sin^2\theta}\mid=\mid\frac{2r^3\cos^2\theta\sin\theta}{r^2}\mid\le 2r\to0$ as $r\to0$.
Hence $f$ is continuous at the origin.