<s>[INST] Differentiability of $f(x,y)= \frac{xy^3}{x^2+y^2}$ for $(x,y)\neq (0,0)$ and $0$ for $(x,y)=(0,0)$. 
Let's consider $f:\mathbb{R}^2\to\mathbb{R}$ with\begin{align*} f:=\begin{cases} \frac{xy^3}{x^2+y^2},&(x,y)\neq (0,0)\\0, & (x,y)=(0,0). \end{cases} \end{align*}
Show that $f$ is differentiable at $(0,0)$.


My approach:
We prove that ${0\choose 0}$ is the (total) derivative at point $(0,0)$.
If $|x|\geq|y|>0$ then
$$
\Big|\frac{xy^3}{(x^2+y^2)\sqrt{x^2+y^2}}\Big|\leq\frac{|x||y|y^2}{(2y^2)|y|\sqrt{2}}=\frac{|x|}{4}.
$$
If $|y|\geq|x|>0$ then
$$
\Big|\frac{xy^3}{(x^2+y^2)\sqrt{x^2+y^2}}\Big|=\frac{|y|}{(\frac{x^2}{y^2}+1)\sqrt{1+\frac{y^2}{x^2}}}\leq |y|.
$$
Keeping those bounds in mind we look at the limit:
\begin{align*} \lim\limits_{{x\choose y}\to{0\choose 0}}\frac{f(x,y)-f(0,0)-{0\choose 0}\left({x\choose y}-{0\choose 0}\right)}{\Vert{x\choose y}-{0\choose 0}\Vert_2 }=\lim\limits_{{x\choose y}\to{0\choose 0}}\frac{\frac{xy^3}{x^2+y^2}}{\Vert{x\choose y}\Vert_2 }=\lim\limits_{{x\choose y}\to{0\choose 0}}\frac{xy^3}{(x^2+y^2)\sqrt{x^2+y^2}},\text{ where } (x,y)\neq (0,0).
\end{align*}
If $x=0$ or $y=0$ then the above limit is $0$. If $|y|\geq|x|>0$ or $|x|\geq|y|>0$ then we use the above upper bounds and see that the limit is again $0$. Hence, $f$ is (total) differentiable at $(0,0)$.

Is this correct? Our tutor told us that when it comes to total differentiability we should rather prove that the partials are continuous. Maybe in this case it was just luck that one could easily see the matrix/derivative.
 [/INST] Noting
$$ \frac{|x|}{\sqrt{x^2+y^2}}\le 1, \frac{y^2}{x^2+y^2}\le1, $$
one has
$$ \bigg|\frac{xy^3}{(x^2+y^2)\sqrt{x^2+y^2}}\bigg|=|y|\frac{|x|}{\sqrt{x^2+y^2}}\cdot\frac{y^2}{x^2+y^2}\le|y|$$
which gives
$$ \lim\limits_{{x\choose y}\to{0\choose 0}}\frac{xy^3}{(x^2+y^2)\sqrt{x^2+y^2}}=0. $$
</s>