Study of differentiablity of function Study the differentiability of the function $f:\mathbb{R}^2\rightarrow \mathbb{R}$
$f(x,y)=\begin{cases} 
      \frac{x^3+y^3}{x^2+\left|y\right|} & (x,y)\ne(0,0) \\
      0 &(x,y)=(0,0) \\
   \end{cases}$
in point $(0,0)$.
So what I did is I calculated the partial derivatives of the function in point $(0,0)$. I got:
$$\frac{∂f}{∂x}\left(0,0\right)=\lim_{t\rightarrow 0}\left(\frac{f\left(t,0\right)-f\left(0,0\right)}{t}\right)=\lim_{t\rightarrow 0}\left(\frac{t^3}{t^3}\right)=1$$and
$$\frac{∂f}{∂y}\left(0,0\right)=\lim_{t\rightarrow 0}\left(\frac{f\left(0,t\right)-f\left(0,0\right)}{t}\right)=\lim_{t\rightarrow 0}\left(\frac{t^3}{t\left|t\right|}\right)=0$$
So he partial derivatives do exist. Next I think I need to calculate:
$$l=\lim_{h\to O_m}\left(\frac{f\left(a+h\right)-f\left(a\right)-<h,\Delta f\left(a\right)>}{\left|\left|h\right|\right|}\right)$$
$$\:l=\lim_{\left(h_1,h_2\right)\rightarrow \left(0,0\right)}\:\left(\frac{f\left(h_1,h_2\right)-f\left(0,0\right)-h_1\cdot 1-h_2\cdot 0}{\sqrt{h_1^2+h_2^2}}\right)$$
$$\:l=\lim_{\left(h_1,h_2\right)\rightarrow \left(0,0\right)}\:\left(\frac{\frac{h_1^3+h_2^3}{h_1^2+\left|h_2\right|}-h_1}{\sqrt{h_1^2+h_2^2}}\right)$$
And this is the point at which I get stuck, since I don't really fully understand what I'm doing. Can anyone help me with this a bit?
 A: To show that $f$ is differentiable at $(0,0)$, we need to show that 
$$
0 = \lim_{\Vert h \Vert \to 0}\left(\frac{f(0+h)-f(0)-\langle h, \Delta f(0)\rangle}{\Vert h \Vert}\right)
$$
with $\Delta f(0)=(\frac{\partial f}{\partial x}(0),\frac{\partial f}{\partial y}(0))$. 
In this case, the answer is no, the function is not differentiable at $(0,0)$. By your calculations, we have
$$
\frac{f(0+h)-f(0)-\langle h, \Delta f(0)\rangle}{\Vert h \Vert} = \frac{\frac{h_1^3+h_2^3}{h_1^2+\vert h_2 \vert}-h_1}{\sqrt{h_1^2+h_2^2}}
$$
Consider now $h=(t,t)$ with $t>0$. Then 
\begin{eqnarray}
\frac{f(0+h)-f(0)-\langle h, \Delta f(0)\rangle}{\Vert h \Vert} & = & \frac{\frac{2t^3}{t^2+t}-t}{t\sqrt{2}}\\
& = & \frac{2t^3-t^3-t^2}{(t^2+t)t\sqrt{2}}\\
& = & \frac{t^2-t}{(t^2+t)\sqrt{2}}\\
& = & \frac{t-1}{(t+1)\sqrt{2}}
\end{eqnarray}
Hence, when $h$ is of the form $h=(t,t)$ with $t>0$, and $\Vert h\Vert\to 0$ (which implies $t\to 0$), we have
$$
\frac{f(0+h)-f(0)-\langle h, \Delta f(0)\rangle}{\Vert h \Vert} = \frac{t-1}{(t+1)\sqrt{2}} \to \frac{-1}{\sqrt{2}} \neq 0
$$
which shows 
$$
\frac{f(0+h)-f(0)-\langle h, \Delta f(0)\rangle}{\Vert h \Vert}
$$
doesn't converge to $0$ when $\Vert h \Vert \to 0$. 
