Let $f(x,y)= \begin{cases} \frac{2x^2y}{x^2+y^2} & if & (x,y) \neq (0,0) \\ 0 & if & (x,y)=(0,0) . \end{cases}$
(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 $t$ aproaches 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? Can anyone help me end the proof of continuity or not continuity , please?
Directional derivative is defined as $$D_\vec{u}f(x,y)=\lim_{h\to0}\dfrac{f(x+ah,y+bh)-f(x,y)}{h} $$ where $\vec{u}=\langle a,b\rangle$ is the unit vector along which rate of change of $f(x,y)$ is calculated.
$$\displaystyle D_\vec{u}f(0,0)=\lim_{h\to0}\dfrac{\dfrac{2(ah)^2(bh)}{(ah)^2+(bh)^2}-0}{h}= \lim_{h\to0}\frac{2a^2b}{a^2+b^2}$$ which exists since $||\vec{u}||=1\implies a^2+b^2\neq0$. So directional derivatives of $f$ exist in any direction at the point $(0,0)$ since limit is independent of how $h$ approaches $0$.
To calculate continuity let $(x,y)\to(0,0)$ along $y=mx$ where $m$ is a constant.
$$\displaystyle\lim_{(x,mx\to(0,0))}\dfrac{2x^2mx}{x^2(1+m^2)}=\lim_{\substack{x\to 0\\\text{along y=mx}}}\dfrac{2mx}{(1+m^2)}=0=f(0,0)$$
So $f$ is continuous at $(0,0)$.