For the function $f(x,y)=|xy|^{p}$ , find the value of $p$ for which $f$ is differentiable at $(0,0)$. MY ATTEMPT:
\begin{equation}
fx(0,0)= \lim_{h\to 0}\frac{f(0+h,0)-f(0,0)}{h}=0
\end{equation}
\begin{equation}
fy(0,0)= \lim_{k\to 0}\frac{f(0,0+k)-f(0,0)}{k}=0
\end{equation}
By Using the definition of differentiability for the functions of two variables:
\begin{equation}
 \lim_{(h,k)\to (0,0)}\frac{f(a+h,b+k)-f(a,b)-hf_x(a,b)-kf_y(a,b))}{{\sqrt{h^2+k^2}}}
\end{equation}
\begin{equation}
\lim_{(h,k)\to (0,0)}\frac{|hk|^p}{\sqrt{h^2+k^2}}=\text{0 if and only if p=2n where n=1,2,3..} 
\end{equation}
so, $p=2n (n=1,2,3\ldots)$
PLEASE CHECK whether I attempted this question correctly or not and if there is any error please give suggestions to resolve it.
Thank you.
 A: We'll try to prove that your function is differentiable if and only if $p >1/2$
For the case $p > 1/2$:
We know that : $\sqrt {{h^2}+{k^2}}\ge \sqrt {2\lvert h \rvert \lvert k \rvert}$, thus :
$$0\le \frac{\lvert hk \rvert ^p}{\sqrt {{h^2}+{k^2}}}\le \frac{\lvert hk \rvert ^p}{{\sqrt2 \lvert hk \rvert ^{1/2}}}=\frac{1}{\sqrt2}{(\lvert hk \rvert) ^{p-1/2}} $$
therefore if $p>1/2$, $$\lim\limits_{(h,k) \to (0,0)}\frac{\lvert hk \rvert ^p}{\sqrt {{h^2}+{k^2}}}=0$$ and the function is differentiable at (0,0)
Now if $p\le1/2$ : let's have a look at the limit of $\frac{\lvert hk \rvert ^p}{\sqrt {{h^2}+{k^2}}}$ along the curves : $k=mh , m\in \mathbb{R}^*$ :
$$\lim\limits_{h \to 0}\frac{\lvert mh^2\rvert ^p}{\sqrt {{h^2}+{m^2h^2}}}=\lim\limits_{h \to 0}\frac{{\lvert m\rvert}^p{\lvert h\rvert}^{2p-1}}{\sqrt {1+m^2}}$$
If $p=1/2$ the limit depends on m  thus there's no limit and the function is not differentiable at $(0,0)$
If $p<1/2$ limit is $\infty$ so the function is also not differentiable at $(0,0)$
