How to deduce the derivative of a function from the formal definition of the derivative? Define $f:\mathbb{R}^2 \rightarrow \mathbb{R}$ by
$$ f{x \choose y} = \left\{  \begin{align}  \frac{xy^2}{\sqrt{x^2+y^2}} ,\,& {x \choose y} \ne \mathbf{0}  \\ 
0 ,\, & {x \choose y} = \mathbf{0}\end{align} \right. $$
Show that $f$ is differentiable at $\mathbf{a}=0$ and determine $Df(\mathbf{0})$.
The solution I am provided with begins by stating:

We use the definition to show that $Df(\mathbf{0})=\begin{align}[ 0 \hspace{0.5 cm} 0] \end{align}$...

I don't understand how to pick $Df(\mathbf{0})=\begin{align}[ 0 \hspace{0.5 cm} 0] \end{align}$ to test against the formal definition of the derivative, since the formal definition says that:
If $A$ is such that:
$\frac{f(\mathbf{a+h}-f(\mathbf{a})-A\mathbf{h})}{||\mathbf{h}||}\rightarrow \mathbf{0}$ as $\mathbf{h} \rightarrow \mathbf{0}$
Then $A$ is the derivative of $f$ at $\mathbf{a}$.
But it does not specify how to pick the $A$. So how did the solution know to pick $A=\begin{align}[ 0 \hspace{0.5 cm} 0] \end{align}$? Was it obvious from the definition of $f$ or did it rely on other properties or theorems?
 A: $A$ is a bounded linear map such that you can write
$$f\left(a+h\right)=f\left(a\right)+A\left(h\right)+\mathcal{R}_{a}\left(h\right)$$
where $\mathcal{R}_{a}$ is a map such that
$$\lim_{h\rightarrow0}\frac{1}{\|h\|}\mathcal{R}_{a}\left(h\right)=0.$$
It is equivalent to say that
$$\lim_{h\rightarrow0}\frac{f\left(a+h\right)-f\left(a\right)-A\left(h\right)}{\|h\|}=0.$$
In particular, write $$\frac{\left(x+k\right)\left(y+\ell\right)^2}{\sqrt{\left(x+k\right)^2+\left(y+\ell\right)^2}}$$
as
$$\frac{xy^2}{\sqrt{x^2+y^2}}+A\left(\left(k,\ell\right)\right)+\mathcal{R}_{\left(x,y\right)}\left(\left(k,\ell\right)\right)$$
with $A$ a bounded linear map in $\left(k,\ell\right)$ and with
$$\lim_{\left(k,\ell\right)\rightarrow\left(0,0\right)}\frac{1}{\|\left(k,\ell\right)\|}\mathcal{R}_{\left(x,y\right)}\left(\left(k,\ell\right)\right)=0.$$
Then, $A_{\left(x,y\right)}$ will be the differential of $f$ at the point $\left(x,y\right)$, noted $\mathrm{d}f\left(\left(x,y\right)\right)$ or $Df\left(\left(x,y\right)\right)$.
A: If $f$ is differentiable at $(0,0)$ the partial derivatives ${\partial f\over\partial x}(0,0)$ and ${\partial f\over\partial y}(0,0)$ do exist, and together they then determine $df(0,0)$. Now
$${\partial f\over\partial x}(0,0)=\lim_{x\to0}{f(x,0)-f(0,0)\over x}=\lim_{x\to0}{0\over x}=0\ ,$$ and similarly  $${\partial f\over\partial y}(0,0)=0\ .$$
Therefore the only way for $f$ to be differentiable at $(0,0)$ would be that $df(0,0)=0$. It remains to test whether in fact
$$\left(\frac{f(\mathbf{a+h})-f(\mathbf{a})-A\mathbf{h}}{||\mathbf{h}||}=\right)\qquad {f(x,y)-f(0,0)-0\over\sqrt{x^2+y^2}}={xy^2\over x^2+y^2}=r\cos\phi\sin^2\phi$$
tends to $0$ when $r\to0$. Since this is obviously the case we can conclude that $f$ is differentiable at $(0,0)$ with $df(0,0)=0$, resp. $=[0\ 0]$.
