Differentiability of a Multi-variable function $\frac{xy}{\sqrt {x^2+y^2}}$. In my book, I found a problem, asked to check the differentiability of the function $f(x,y)$ at $(0,0)$ where the function f$(x,y)$ is defined as follows

$$
f(x,y) =
\begin{cases}
\large\frac{xy}{\sqrt {x^2+y^2}},  & \text{if $x^2+y^2 \neq (0,0)$} \\
0, & \text{if $x^2+y^2 = (0,0)$}
\end{cases}
$$

Now in the book, they give a solution using alternative definition of differentiability at a point $(0,0)$
and there they have approached to a contradiction that $\frac{1}{2} = 0$ 
So, they proved that the function is not differentiable at all.
But when I tried to solve the problem by proving that the $f_x$ and $f_y$ exist at $(0,0)$ and the function is continuous at that point too. So basically, I prove that the given function is differentiable at that very point.
But the given solution in my book makes me confuse.
I don't understand what actually happen, Did I do something wrong if I did please make me clear? Because at the same time these two solution is not possible for this function.
And another thing can I use "sufficient condition for differentiability" for this function?
Pardon me if I did any wrong.
Thank you.
 A: Let me try to show this using the very definition of differentiability. Suppose the function $f$ is differentiable at $(0,0)$. Then there is a matrix $D = (a,b)$ (the derivative of $f$ at zero) such that
$$ 0 = \lim_{(x,y)\to (0,0)} \frac{|f(x,y) - f(0,0) - D \cdot (x,y)^T|}{\|(x,y) - (0,0)\|} = \lim_{(x,y)\to(0,0)} \frac{1}{\sqrt{x^2 + y^2}} \left| \frac{xy}{\sqrt{x^2+y^2}} -ax -by \right|. $$
Let us take two particular ways how to approach zero, say $(x,y) = (h,h)$ as $h \to 0_\pm$ (left/right). Then we have
$$ 0 = \lim_{h \to 0} \frac{1}{\sqrt{2h^2}} \left| \frac{|h|}{\sqrt{2}} - (a+b)h \right| = \frac{1}{\sqrt{2}} \lim_{h\to 0} \left| \frac{1}{\sqrt{2}} - (a+b)\mathrm{sign}(h) \right| = \frac{1}{\sqrt{2}} \left| \frac{1}{\sqrt{2}} \mp (a+b) \right|. $$
So, equalities $a+b = \frac{1}{\sqrt{2}}$ and $a+b = - \frac{1}{\sqrt{2}}$ would have to hold at the same time. This is clearly a contradiction.
A: As in the other answer, there is no need to consider a general linear map. If at all $f$ is differentiable, the associated linear map must be the zero map. Let me explain why.
We have $$\frac{\partial f}{\partial x}\vert_{(0,0)} = 0 \quad\text{and} \quad \frac{\partial f}{\partial y}\vert_{(0,0)} = 0$$
by definition of partial derivatives. Suppose $f$ is differentiable at $(0,0)$. The associated linear map $Df_{(0,0)}:\mathbb R^2\to\mathbb R$ satisfies $Df_{(0,0)}(e_1) = D_{e_1}f(0,0) = 0$ and $Df_{(0,0)}(e_2) = D_{e_2}f(0,0) = 0$. By linearity, $Df_{(0,0)}(h,k) = 0$ for all $(h,k)\in \mathbb R^2$, so $Df_{(0,0)} \equiv 0$. Now, assuming $Df_{(0,0)} \equiv 0$, it is easier to reach a contradiction - in particular by evaluating the limit $$\lim_{(x,y)\to (0,0)} \frac{xy}{\sqrt{x^2+y^2}}\cdot\frac{1}{\sqrt{x^2+y^2}} = \lim_{(x,y)\to (0,0)} \frac{xy}{x^2+y^2}$$
along $y=x$, to get $\frac12\ne 0$.
