Let be $f:\mathbb{R}^2\to\mathbb{R}$ where$$f(x,y):=\begin{cases} \frac{x^3+y^3}{x^2+y^2},&(x,y)\neq (0,0)\\0,&(x,y)=(0,0).\end{cases}$$Show that $f$ is not differentiable at $(0,0)$.
My approach:
If $f$ was differentiable at $(0,0)$, by definition there must exist a unique matrix $A=(a,b)$ such that\begin{align*}&\lim\limits_{{x\choose y}\to{0\choose 0}}\frac{\left|\frac{x^3+y^3}{x^2+y^2}-0-A{x\choose y}\right|}{\Vert {x\choose y}\Vert_2}=0.\end{align*}We consider the sequence $\left(\frac{1}{n},0\right)$ which implies that\begin{align*}&\lim\limits_{n\to\infty}\frac{\left|\frac{\frac{1}{n^3}+0^3}{\frac{1}{n^2}+0^2}-0-A{\frac{1}{n}\choose 0}\right|}{\frac{1}{n}}=1-a\implies a=1.\end{align*}Similarly, $\left(0,\frac{1}{n}\right)$ implies\begin{align*}&\lim\limits_{n\to\infty}\frac{\left|\frac{\frac{1}{n^3}+0^3}{\frac{1}{n^2}+0^2}-0-A{0\choose \frac{1}{n}}\right|}{\frac{1}{n}}=1-b\implies b=1.\end{align*}Finally, we look at the sequence $\left(\frac{1}{n},\frac{1}{n}\right)$ and see that
\begin{align*}&\lim\limits_{n\to\infty}\frac{\left|\frac{\frac{1}{n^3}+\frac{1}{n^3}}{\frac{1}{n^2}+\frac{1}{n^2}}-0-A{\frac{1}{n}\choose \frac{1}{n}}\right|}{\sqrt{2}\frac{1}{n}}=\frac{1}{\sqrt{2}}(1-a-b).\end{align*}As we have already shown that $a=b=1$ it is not possible that the last limit attains $0$. So there doesn't exists a unique matrix $A$.
This apporach feels a bit lengthy but I was not able to produce a contradiction only by using one sequence. Maybe someone else has a swifter approach or can give me some feedback on this.