Function Differentiable at the origin 
Prove $h(x,y) = \big| xy \big|$ is differentiable at the origin but not of $C^1$ in any neighborhood of $0$.

My intuition is that $|x|$ is not differentiable at $0$ since it bends at the origin so it should be the same thing for function $h$, which is not the case here. I saw an approach of the following: 
By Young's inequality,
$$
2 |xy| \leq x^2+y^2
$$
Therefore
$$
\frac{|xy|}{\sqrt{x^2+y^2}} \leq \frac{1}{2} \sqrt{x^2+y^2}
$$
and hence $Df(0,0)=0$. But I do not know how to show that it is not differentiable at the origin by definition simply. To show it is not $C^1$, i have to show it is a differentiable function whose derivative is continuous but I am not sure what exactly the derivative of ths absolute value is.
 A: You have already shown that $f$ is differentiable at the origin. 
Suppose that $y_0>0$. As you say, if you test differentiability at the point $(0,y_0)$ you are looking for a linear map $L:\mathbb R^2\to\mathbb R$ such that 
$$\tag1
\frac{|h_1(y_0+h_2)|-0-L(h_1,h_2)}{\sqrt{h_1^2+h_2^2}}\xrightarrow[h\to(0,0)]{}0.
$$
Since $L$ is linear, it is of the form $L(h_1,h_2)=\alpha h_1+\beta h_2$. 
Now $(1)$ looks like $$\tag2
\frac{|h_1(y_0+h_2)|-\alpha h_1-\beta h_2}{\sqrt{h_1^2+h_2^2}}.
$$
If $h_1>0$ and $h_2=0$, we get
$$\tag3
\frac{h_1y_0-\alpha h_1}{|h_1|}=y_0-\alpha.
$$
If instead we take $h_1<0$ and $h_2=0$, we get 
$$\tag4
\frac{-h_1y_0-\alpha h_1}{|h_1|}=y_0+\alpha.
$$
From $(3)$ and $(4)$ we get that $\alpha=0$. Now $(1)$ became
$$\tag5
\frac{|h_1(y_0+h_2)|-\beta h_2}{\sqrt{h_1^2+h_2^2}}.
$$
Now suppose that $h_1>0$ and $h_2=h_1$. Then $(5)$ becomes 
$$\tag6
\frac{h_1y_0+h_1^2-\beta h_1}{\sqrt{2}\,|h_1|}\to \frac{y_0-\beta}{\sqrt2}.
$$
If instead we take $h_1>0$ and $h_2=mh_1$, now we have 
$$\tag7
\frac{h_1y_0+mh_1^2-m\beta h_1}{\sqrt{m^2+1}\,|h_1|}\to \frac{y_0-m\beta}{\sqrt{m^2+1}}.
$$
As we are free to choose $m$, the limits in $(6)$ and $(7)$ are incompatible, thus showing that $(1)$ is impossible, and so $f$ is not differentiable at $(0,y_0)$. 
A: You have shown that $h$ is differentiable at the origin, with $Dh(0,0)=0$.
On the other hand, let $U$ be an open disc around $(0,0)$. Then $U$ contains a point $(\xi,0)$ with $\xi>0$. I claim that $h$ is not differentiable at $(\xi,0)$; hence $h$ is not $C^1$ in $U$.
Proof. Near $(\xi,0)$ the function $h$ is given by $h(x,y)=x\,|y|$. It follows that the partial derivative $${\partial h\over\partial y}(\xi,0)$$
does not exist. But this would be necessary for the existence of $Dh(\xi,0)$.
