Cfrgtkky

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 $t$ 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?

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 trought $(0,0)$ which gives me another value distinct to $0$. :(

Can anyone help
me end the proof of continuity or not continuity , please?

Using polar coordinates,
$midfrac{2x^2y}{x^2+y^2}mid=midfrac{2r^2cos^2theta rsintheta}{r^2cos^2theta +r^2sin^2theta}mid=midfrac{2r^3cos^2thetasintheta}{r^2}midle 2rto0$ as $rto0$.

Hence $f$ is continuous at the origin.

Using AM-GM inequality, x^2+y^2 >= 2xy. 1/(x^2+y^2) =< 1/2xy.

So |2x^2y/(x^2+y^2)| =< |(2x^2y/2xy = x )| which approaches to zero as x approaches to zero.

Hence it is continuous at (0,0).