<s>[INST] totally differentiable function $\frac{x^3}{(x^2+y^2)}$ - check my proof Consider the function: $f: \mathbb{R^2} \to \mathbb{R}$ given by $f(x,y)=\begin{cases} \frac{x^3}{x^2+y^2} & (x,y) \neq (0,0) \\ 0 & (x,y)=(0,0) \end{cases}$ 
I want to check if it is totally differentiable at ${0 \choose 0}$. By definition this means that there must exists a linear transformation $A$ such that:
$lim_{{x \choose y} \to {0 \choose 0}} \frac{f{x \choose y}-f{0 \choose 0}-A({x \choose y}-{0 \choose 0})}{\Vert {x \choose y} - {0 \choose 0} \Vert} = 0 $, where $\Vert \cdot \Vert$ is an arbitrary norm.
My proof:
I assume that $f$ is totally differentiable so that $A$ must consist of the partial derivatives. The partial derivative at point ${0 \choose 0}$ with respect to $x$ is $D_1f{0 \choose 0}=1$ and with respect to $y$: $D_2f{0 \choose 0}=0$. Let $A:= \left(1~~0\right)$. Plugging this into $lim_{{x \choose y} \to {0 \choose 0}} \frac{f{x \choose y}-f{0 \choose 0}-A({x \choose y}-{0 \choose 0})}{\Vert {x \choose y} - {0 \choose 0} \Vert}$ yields: $lim_{{x \choose y} \to {0 \choose 0}} \frac{f{x \choose y}-\left(1~~0\right){x \choose y}}{\Vert {x \choose y}\Vert}=lim_{{x \choose y} \to {0 \choose 0}} \frac{-xy^2}{(x^2+y^2)\Vert {x \choose y}\Vert}$. At this point, I am not sure how to continue. A first approach might be:
1.) As in $\mathbb{R^2}$ norms are equivalent it holds that $C \cdot \Vert {x \choose y} \Vert_{\infty} \geq \Vert {x \choose y} \Vert $ where $C >0$ and $\Vert {x \choose y} \Vert_{\infty}$ is the maximum norm. If I now assume that $x\leq y<0$, I can conclude: $lim_{{x \choose y} \to {0 \choose 0}} \frac{-xy^2}{(x^2+y^2)\Vert {x \choose y}\Vert}\geq lim_{{x \choose y} \to {0 \choose 0}} \frac{-xy^2}{C(x^2+y^2) \vert y\vert} = lim_{{x \choose y} \to {0 \choose 0}} \frac{-x\vert y \vert}{C(x^2+y^2)}$. If $f$ were totally differentiable then we know that for an arbitrary null sequence $lim_{{x \choose y} \to {0 \choose 0}} \frac{-x\vert y \vert}{C(x^2+y^2)}$ must be $0$. However, if I plug in the null sequence ${\frac{-1}{n} \choose \frac{-1}{n}}$ the limit is $\frac{1}{2C}> 0$. Hence, $f$ is not totally differentiable at ${0 \choose 0}$.
A second approach could be:
2.) If I plug the null sequence ${\frac{-1}{n} \choose \frac{1}{n}}$ directly into $lim_{{x \choose y} \to {0 \choose 0}} \frac{f{x \choose y}-f{0 \choose 0}-A({x \choose y}-{0 \choose 0})}{\Vert {x \choose y} - {0 \choose 0} \Vert} $ then it should equal $0$ if $f$ was totally differentiable. However, rearrangeing yields: $lim_{{x \choose y} \to {0 \choose 0}} \frac{-(\frac{-1}{n})(\frac{1}{n})^2}{((\frac{-1}{n})^2+(\frac{1}{n})^2)\Vert {\frac{-1}{n} \choose \frac{1}{n}}\Vert}= lim_{{x \choose y} \to {0 \choose 0}} \frac{\frac{1}{n^3}}{\frac{2}{n^2}\cdot \frac{1}{n}\Vert {-1 \choose 1}\Vert}=lim_{{x \choose y} \to {0 \choose 0}} \frac{1}{2\cdot\Vert {-1 \choose 1}\Vert} > 0$. Hence, $f$ is not totally differentiable at ${0 \choose 0}$.
May be those approaches are not that straightforward but as I am currently learning multivariable calculus I am interested in wether the proves are correct or not. Any comments or suggestions are appreciated. 
best regards Philipp
 [/INST] What you did is correct.
Maybe a simpler way would be to notice that for $x=y=t>0$
$$\frac{-xy^2}{(x^2+y^2)\Vert {x \choose y}\Vert}=-\frac{1}{2\sqrt 2}$$
and that consequently, above function can’t have $0$ for limit at $(0,0)$.
</s>