For which $u$ does the derivative $f'(u,0)$ of $f(x,y)=x^3/(x^2+y^2)$ exist? Let $f: \mathbb{R}^2\to \mathbb{R}$ be defined by setting
$$f(x,y)=\begin{cases}\frac{x^3}{x^2+y^2} & (x,y)\neq (0,0)\\0 & (x,y)=(0,0)\end{cases}$$
(a) For which vectors $u\ne 0$ does $f'(0; u)$ exist? Evaluate it when it
exists.
(b) Do $D_1f$ and $D_2f$ exist at $0$?
(c) Is $f$ differentiable at $0$?
(d) Is $f$ continuous at $0$?
I have thought a lot about this problem: 
For (a), be $u\neq 0, u:=(h,k)$, so $\lim_{t\to 0}\frac{f(0+tu)-f(0,0)}{t}=\lim_{t\to 0}\frac{f(th,tk)}{t}=\lim_{t\to 0}\frac{(th)^3}{(th)^2+(tk)^2}\frac{1}{t}=\frac{h^3}{h^2+k^2}$, then $f'(0,u)$ exists for all $u\neq 0$.
For (b), $\lim_{t\to 0}\frac{f(0+te_1)-f(0)}{t}=\lim_{t\to 0}\frac{f(t,0)}{t}=\lim_{t\to 0}\frac{t^3}{t^3}=\lim_{t\to 0}1=1$ so $D_1f(0,0)=1$ and $\lim_{t\to 0}\frac{f(0+te_2)-f(0)}{t}=\lim_{t\to 0}0=0$ so $D_2f(0,0)=0$
I am having problems solving (c) and (d), could someone help me please? Thank you very much.
 A: Since, $|x|\le \|(x,y)\|$ we have, $$|f(x,y)| =\frac{|x^3|}{x^2+y^2} =\frac{|x^3|}{\|(x,y)\|^2}\le \frac{\|(x,y)\|^3}{\|(x,y)\|^2}=\|(x,y)\|\to0$$
This prove the continuity of $f$.
For the differentiability we have $$\frac{f(h,0)-f(0,0)}{h}= \frac{h^3}{h^3} = 1$$
and 
$$\partial_x f(0,0)=\frac{f(h,h)-f(0,0)}{h}= \frac{h^3}{2h^3} = \frac{1}{2}$$
whereas
$$\partial_y f(0,0)=\frac{f(0,h)-f(0,0)}{h}= 0$$
Then f is not differentaible at (0,0). 

Indeed we have $\nabla f(0,0) = (1,0)$
  But $$\lim_{(h, k)\to (0,0)} \frac{f(h,k)-f(0,0)}{\|(h,k)\|}-\nabla f(0,0)\cdot (h,k)\neq 0$$
  Because on the diagonal direction we have:
  $$\lim_{(h, h)\to (0,0)} \frac{f(h,h)-f(0,0)}{\|(h,h)\|}-\nabla f(0,0)\cdot (h,h)= \lim_{h\to 0}\frac{h^3}{\sqrt{2}|h|h^2}-1\cdot h+0\cdot h) = ±\frac{1}{\sqrt{2}} \neq 0. $$

A: Is my solution for (c) ok?
(c) Assume that $f$ is differentiable at $\mathbf{0}$.
Let $\mathbf{u}=(1,1)$.
$f'(\mathbf{0};\mathbf{u})=Df(\mathbf{0})\cdot\mathbf{u}=\begin{bmatrix} D_1f(\mathbf{0})&D_2f(\mathbf{0})\end{bmatrix}\cdot\begin{bmatrix}1\\1\end{bmatrix}=\begin{bmatrix}1&0\end{bmatrix}\cdot\begin{bmatrix}1\\1\end{bmatrix}=1$.
On the other hand, from (a), $f'(\mathbf{0};\mathbf{u})=\frac{1}{2}$.
This is a contradiction.
So, $f$ is not differentiable at $\mathbf{0}$.
(d) Let $\mathbf{h}=(h_1,h_2)$.
$|f(h_1,h_2)-f(0,0)|=|\frac{h_1^3}{h_1^2+h_2^2}|\leq|h_1|\frac{|h_1^2|}{|h_1^2+h_2^2|}\leq|h_1|\frac{|h_1^2+h_2^2|}{|h_1^2+h_2^2|}\leq|h_1|\leq\max\{|h_1|,|h_2|\}=|\mathbf{h}|.$
So, $f$ is continuous at $\mathbf{0}$.
