Find $p$ for $f(x,y)$ to be differentiable. 
Find $p$ such that $f(x,y)$ is differentiable at $(x,y)=(0,0)$
$$f(x,y)=\begin{cases} |xy|^p\ \ ,xy\neq0\\0\ \ \ \ \ \ \ \  ,xy=0\end{cases}$$

I got, $$f_{x}(0,0)=\lim_{h\rightarrow0}\dfrac{f(0+h,0)-f(0,0)}{h}=0$$
Also,$$f_{y}(0,0)=\lim_{h\rightarrow0}\dfrac{f(0,0+h)-f(0,0)}{h}=0$$
For drivative I'll use following,
$$\begin{align}\triangle z&=dz+\epsilon_1\triangle x+\epsilon_2\triangle y\\f(\triangle x,\triangle y)-f(0,0)&=0+0+\epsilon_1\triangle x+\epsilon_2\triangle y\\|\triangle x\cdot\triangle y|^p &=\epsilon_1\triangle x+\epsilon_2 \triangle y\end{align}$$
Comparing both sides,$$\begin{align}\epsilon_1&=\dfrac{|\triangle x\cdot\triangle y|^p}{\triangle x}\\\epsilon_2&=0\end{align}$$
Now, $$\begin{align}\lim_{(\triangle x,\triangle y)\rightarrow(0,0)}\epsilon_1&=0\\\lim_{(\triangle x,\triangle y)\rightarrow(0,0)}\dfrac{|\triangle x\cdot\triangle y|^p}{\triangle x}&=0\end{align}$$
As$$0\leq\dfrac{|\triangle x\cdot\triangle y|^p}{|\triangle x|}=|\triangle x|^{p-1}\cdot|\triangle y|^p$$
Then, $$p\geq1$$
But answer is $p\geq\dfrac{1}{2}$. Please tell me what I'm missing.
 A: $\newcommand{\dx}{\Delta x}$
$\newcommand{\dy}{\Delta y}$
See, I understand till the point $|\dx \cdot \dy|^p = \epsilon_1 \dx + \epsilon_2 \dy$.
After that, there are multiple possibilities for $\epsilon_1 , \epsilon_2$ so that the equations can be satisfied. If we can find one such combination for which $\epsilon_1,\epsilon_2$ go to zero, then we are done. 
Your  problem is that you found only one possible combination (namely $\epsilon_1 = \frac{|\dx \cdot \dy|^p}{|\dx|}$ and $\epsilon_2 = 0$). Another combination exists, for example $\epsilon_1 = 0$ and $\epsilon_2 = \frac{|\dy \cdot \dx|^p}{|\dy|}$. You should try all combinations before confirming non-differentiability, but you tried only one combination, which workd only when $p \geq 1$. Some other combination of $\epsilon_1,\epsilon_2$ will work for the other $\frac 12 < p \leq 1$.

Here's what works : take $\epsilon_1 = \epsilon_2 = \frac{|\dx \cdot \dy|^p}{\dx + \dy}$ if $\dx + \dy \neq 0$, and both equal to $0$ otherwise. The claim is that for $p > \frac 12$, this works. I leave you to see that.

However, at $p= \frac 12$, the function is not differentiable. To see this, take the path $x = y$ travelling to $(0,0)$ to see that $\frac{|xy|^{0.5}}{\sqrt{x^2+y^2}} = \frac{1}{\sqrt 2}$, so the limit does not equal zero. For any $p < \frac 12$, this same path limit is infinite, so the limit won't exist.
A: By the definition of differentiability for functions $f:\mathbb{R}^2\to\mathbb{R}$ the limit$^\dagger$ $$\lim_{\mathbf{h}\to 0}\frac{|f(x+h,y+h)-f(x,y)-( f_x(x,y), f_y(x,y) ) \cdot \mathbf{h}|}{||\mathbf{h}||}$$ must exist and must equal $0$ at $(x,y)$. From your calculations, let's plug in what we know
$$\lim_{\mathbf{h}\to 0}\frac{||h_1h_2|^p-0-\mathbf{0} \cdot \mathbf{h}|}{||\mathbf{h}||} = \lim_{\mathbf{h}\to 0}\frac{|h_1h_2|^p}{\sqrt{h_1^2+h_2^2}} = 0$$
Converting to polar coordinates
$$\lim_{r\to 0^+} \frac{r^{2p}|\sin\theta\cos\theta|^p}{r} = \lim_{r\to 0^+} r^{2p-1}|\sin\theta\cos\theta|^p = 0$$
In order for squeeze theorem to hold, $2p-1 > 0$, thus $p > \frac{1}{2}$
$^\dagger$EDIT: Technically inside the limit the role of the gradient in the true definition is played by some arbitrary linear operator, and the condition for differentiability is if such an operator exists that makes the limit equal to $0$, then the function is said to be differentiable at $(x,y)$. So we would have to handle the case $p=\frac{1}{2}$ separately because a different linear operator besides the $\mathbf{0}$ operator might make the limit exist and equal $0$. But I don't think that is the case here.
