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Let f be a function that satysfies : $$f: \mathbb{R}^2 \rightarrow \mathbb{R} , (x,y) \mapsto \begin{cases} 0 & \text{for } (x,y)=(0,0) \\ \frac{x^3}{x^2+y^2} & \text{for } (x,y) \neq (0,0) \end{cases} $$ I want to show that f has directional derivative at point $(0,0)$ at every direction. So, let $v=(v_1,v_2) \in R^2, (v_1,v_2) \neq (0,0)$. f has directional derivative in $(0,0)$ if and only if following limit exist : $$\lim_{t->0}\frac{f(a+tv)-f(a)}{t}=\lim_{t->0}\frac{f(tv_1,tv_2)}{t}=\lim_{t->0}\frac{t^3v_1^3}{t(t^2v_1^2+t^2v_2^2)}=\frac{v_1^3}{v_1^2+v_2^2} $$.

So i show that above limit exist and it depends of choosing vector v. So the directional direvative exist at point $(0,0)$ in every direction. The main question is :

Am i thinking correcly ?

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Yes, you are right. The directional derivative exists at $(0,0)$ in every dirction $v=(v_1,v_2)$ and is $=\frac{v_1^3}{v_1^2+v_2^2}.$

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