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.
 A: It looks correct, but it's easier to see that, if $x=\rho\cos(\theta)$ and $y=\rho\sin(\theta)$, then$$\left|\frac{xy^3}{(x^2+y^2)^{3/2}}\right|=\rho|\cos(\theta)\sin^3(\theta)|\leqslant\rho=\sqrt{x^2+y^2}.$$Therefore,$$\lim_{(x,y)\to(0,0)}\frac{xy^3}{(x^2+y^2)^{3/2}}=0.$$
A: Your reasoning looks correct but it also looks a bit nonstandard. Let me give you another point of view which is probably more natural when dealing with these sorts of problems.
First, try to convince yourself that if the derivative of a multivariate function $f$ exists at $x$ (let us denote it by $Df(x)$), it satisfies:
$$Df(x)v = \lim_{t\to 0}\frac{f(x+tv)-f(x)}{t}$$
for every $v$.
In your case, the above limit at $x=0$ becomes:
$$\lim_{t\to 0}\frac{f(tv)-f(0)}{t} = \lim_{t\to 0}\frac{t^{4}hk^{3}}{t^{3}(h^{2}+k^{2})} = 0$$
where I wrote $v = (h,k) \in \mathbb{R}^{2}$. Hence, the natural candidate for $Df(0)v$ is zero. The problem then reduces to check whether the limit:
$$\lim_{(h,k)\to (0,0)}\frac{hk^{3}}{(h^{2}+k^{2})^{\frac{3}{2}}} \tag{1}\label{1}$$
exists. Take $h = r\cos\theta$ and $k = r\sin\theta$. Then:
$$\frac{hk^{3}}{(h^{2}+k^{2})^{\frac{3}{2}}} = r\cos\theta\sin^{3}\theta \tag{2} \label{2}$$
so that when $(h,k) \to (0,0)$, we have $r \to 0$ and (\ref{1}) goes to zero because of (\ref{2}). This limit independs of the path taken, because it does not depend on $\theta$. Thus, the derivative of $f$ at zero exists and equals to zero.
A: 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
$$ \lim\limits_{{x\choose y}\to{0\choose 0}}\frac{xy^3}{(x^2+y^2)\sqrt{x^2+y^2}}=0. $$
