To have differentiabilility, do we have to obtain $\epsilon_1=0$ as $h, k \to 0$ and also $\epsilon_2=0$ as $h, k \to 0$. A function $f(x,y)$ is differentiable  at $(a,b)$ if $f_x(a,b)$ and $f_y(a,b)$ exist and satisfies equation 
$f(a+h,b+k)=f(a,b)+hf_x(a,b)+k_y(a,b)+h\epsilon_1+k\epsilon_2$
in which $\epsilon_1, \epsilon_2 \to 0$ as both $h,k \to 0$. 
According to this definition, $f(x,y)=\sqrt{xy}$ at $(0,0)$  is not differentiable. As a result  we have  $\sqrt{xy}=h\epsilon_1 + k\epsilon_2 $ by applying the differentiability equation above. When $h=k$, 
$\lim_{k\to 0}(\epsilon_1, \epsilon_2)=1 \;\text{or } -1$ 
so $\epsilon_1, \epsilon_2$ cannot be $0$. Hence it is not differentiable. Here we have another example which is differentiable at $(0,0)$
\begin{align}
f(x,y)=
\begin{cases}
(x^2+y^2)\sin\frac{1}{\sqrt{x^2+y^2}}&, (x,y)\neq (0,0),\\
0,&   (x,y)=(0,0) 
\end{cases}
\end{align}
 In this example we have $\lim_{h \to 0} \epsilon_1 +\epsilon_2=0$ when $h=k$. I am confused here. This example is differentiable at $(0,0)$. To have differentiability, do we have to obtain $\epsilon_1=0$ as $h, k \to 0$ and also $\epsilon_2=0$ as $h, k \to 0$. 
Thank you for your help
 A: I'm not really sure what your question it, but let me restate the facts as you know them, inserting some explanations.

A function $f(x,y)$ is differentiable at $(a,b)$ if $f_x(a,b)$ and $f_y(a,b)$ exist, and there exist functions $\epsilon_1(h,k)$ and $\epsilon_2(h,k)$ such that
  $$
    f(a+h,b+k)=f(a,b)+hf_x(a,b)+k_y(a,b)+h\epsilon_1+k\epsilon_2
$$
  and 
  $$
    \lim_{(h,k) \to (0,0)} \epsilon_1(h,k) = \lim_{(h,k) \to (0,0)} \epsilon_2(h,k) = 0
$$

Fact 1. The function $f(x,y) = \sqrt{xy}$ is not differentiable at $(0,0)$.
Proof.  From the definition of partial derivative we have $f_x(0,0) = f_y(0,0)=0$.  If $f$ is differentiable at $(0,0)$, the exists functions $\epsilon_1$ and $\epsilon_2$ as in the definition, satisfying
$$
    \sqrt{hk} = h \epsilon_1(h,k) + k \epsilon_2 (h,k)
$$
Setting $h=k$ gives
$$
    |h| = h(\epsilon_1(h,h) + \epsilon_2(h,h))
    \implies \epsilon_1(h,h) + \epsilon_2(h,h) = \frac{|h|}{h}
$$ 
As $h\to 0^+$, the expression on the left tends to $0+0 = 0$, while the one on the right is $1$.  This is a contradiction.  $\Box$
Fact 2. The function $f(x,y) = (x^2+y^2)\sin \frac{1}{\sqrt{x^2+y^2}}$ is differentiable at $(0,0)$.
Proof.  Again from the definition we can see $f_x(0,0) = f_y(0,0) = 0$.  We must find functions $\epsilon_1(h,k)$ and $\epsilon_2(h,k)$ such that
$$
    (h^2+k^2)\sin \frac{1}{\sqrt{h^2+k^2}} = h\epsilon_1(h,k) + k\epsilon_2(h,k)
\tag{$*$}
$$
Let's try:
\begin{align*}
    \epsilon_1(h,k) &= h\sin \frac{1}{\sqrt{h^2+k^2}} \\
    \epsilon_2(h,k) &= k\sin \frac{1}{\sqrt{h^2+k^2}} \\
\end{align*}
You can see this satisfies ($*$).  It remains to show the epsilons tend to zero.
But
\begin{align*}
    \left|\epsilon_1(h,k)\right| &\leq |h| \cdot 1 = |h| \to 0 \\
    \left|\epsilon_2(h,k)\right| &\leq |k| \cdot 1 = |k| \to 0
\end{align*}
as $(h,k) \to 0$.  So 
$$
    \lim_{(h,k) \to (0,0)} \epsilon_1(h,k) = \lim_{(h,k) \to (0,0)} \epsilon_2(h,k) = 0
\qquad\Box
$$
