$$f(x,y)=\begin{cases}\dfrac{y^3}{x^2+y^2} &(x,y) \neq \ \mathbb{(0,0)}\\ 0 & (x,y)=(0,0) \\ \end{cases}$$
Evaluate $f_x(0,0)$ and $f_y(0,0)$ and $D_\overrightarrow{u}f(0,0)$
I tried directly taking the derivative to no avail (obviously) so then I tried to use the definition of partial derivative which also left me without a correct solution. Also I have proved that it is continuous (Sertoz Theorem) but how would I prove that it is also differentiable at $(0,0)$?
On
Using polar coordinates $$\begin{gather}x = \rho\cos{\varphi}, \\ y= \rho\sin{\varphi}.\end{gather}$$ we have $${\frac{f(x,\,y) - f(0,\,0)}{\sqrt{x^2+y^2}}} ={\frac{f(\rho\cos{\varphi},\,\rho\sin{\varphi}) - 0}{\rho}} = {\frac{\rho^3\sin^3{\varphi}}{\rho^2\rho}} = \sin^3 \varphi$$
Since this can take arbitrary values when $\varphi$ changes, it means that the limit do not exists and so $Df(0,\,0)$ do not exists, i.e $f$ not differentiable in $(0,0)$.
$$f_x(0,0) = \lim_{t\rightarrow 0} \frac{f(t,0) - f(0,0)}{t} = 0, \lim_{t\rightarrow 0} \frac{f(0,t) - f(0,0)}{t} = 1\ \ $$
Next, I will show that $f$ is not differentiable at $(0,0)$.
Let $\vec{u}$ be a unit vector in $\mathbb{R}^2$ and let $\vec{u} = u_1e_1 + u_2e_2$, where $e_1 = (1, 0)$ and $e_2=(0,1)$ are the canonical basis vectors. Then, $$ D_{\vec{u}}f(0,0) = \lim_{t\rightarrow 0} \frac{f((0,0) + t(u_1, u_2))-f(0,0)}{t}\\ = \lim_{t\rightarrow 0}\frac{f(tu_1, tu_2)}{t} \\ = \lim_{t\rightarrow 0}\frac{t^3u_2^3}{t(t^2u_1^2 + t^2u_2^2)} = u_2^3 $$ If $f$ was differentiable at $(0,0)$, then $$D_{\vec{u}}f(0,0) = \nabla f(0,0) \cdot \vec{u} \\ = f_x(0,0)u_1 + f_y(0,0)u_2 = u_2 $$ However, this contradicts what we have above, as $\vec{u}$ can be chosen so that $u_2\ne u_2^3$.