Proof that $f$ is differentiable in $0,0$ where $f(x,y) = \frac{\sqrt{1+xy}-1}{y}$ 
Proof that $f$ is differentable in $0,0$ where $f(x,y) =   \begin{cases} \frac{\sqrt{1+xy}-1}{y} \mbox{ when $y \neq 0$ } \\ \frac{x}{2} \mbox{ when $y = 0$ } \end{cases} $  $f$ is defined on set $A = \left\{ (x,y) \in \mathbb R^2 : xy>-1 \right\}$

My approach
On the beginning I want to show that $f$ is continous, then I can compute partial derivatives... 
Let take any two sequences $x_n \rightarrow x$ and $y_n \rightarrow 0$ then we have
$$ \frac{\sqrt{1+xy}-1}{y}  = \frac{x}{1+\sqrt{1+xy}}  $$
and now
$$ ?\le \frac{x_n}{1+\sqrt{1+x_n y_n}} \le \frac{x_n}{1+1} \rightarrow \frac{x}{2} $$
but how can I bound that from left side?
 A: Hint: $\dfrac{x}{1+\sqrt {1+xy}}$ is the quotient of two functions that are differentiable at $(0,0),$ with the denominator $\ne 0$ at $(0,0).$
A: Let's see that 
$$
\lim_{h\rightarrow0}\frac{f(h,0)-f(0,0)}{h}=\frac{1}{2}
$$
and
$$
\lim_{h\rightarrow0}\frac{f(0,h)-f(0,0)}{h}=0
$$
so if the partial derrivatives exist in $(0,0)$ point they have to be equal to these values.
Now we calculate partial derrivatives and get that
$$
\frac{\partial f}{\partial x} =   \left\{
\begin{array}{ll}
      \frac{1}{2 \sqrt{x y+1}} & y\neq0 \\
      \frac{1}{2} & y=0 \\
\end{array} 
 \right.
\\
\frac{\partial f}{\partial y} =   \left\{
\begin{array}{ll}
      \frac{2 \sqrt{x y+1}-2-x y}{2 y^2 \sqrt{x y+1}} & y\neq0 \\
      0 & y=0 \\
\end{array} 
 \right.
$$
so you see that
$$
\lim_{(x,y)\rightarrow(0,0)}\frac{\partial f}{\partial x} =\frac{1}{2}\\
\lim_{(x,y)\rightarrow(0,0)}\frac{\partial f}{\partial y} =0
$$
what one can calculate using the Taylor series expansion of given functions.
Now you can state that your function is differentiable in $(0,0)$ as it has continuous partial derrivatives in this point. What more, you can even state that it has a continuous differential in this point as we know that in this case
$$
Df(\mathbb{x})=\left[\frac{\partial f}{\partial x}(\mathbb{x}) , \frac{\partial f}{\partial y} (\mathbb{x})\right]
$$
and as we know $Df_1$ and $Df_2$ are continuous in $(0,0)$ so $Df$ is continuous in $(0,0)$
