Calculus exercise - differentiability and $C^1$ functions Show that the function $f(x,y) = |xy|$ is differentiable at $\mathbf{0}$, but is not of $C^1$ in any neighbourhood of $\mathbf{0}$
So a function is differentiable at $\mathbf{0}$ if $\lim\limits_{\mathbf{h} \to 0} \dfrac{f(\mathbf{0} + \mathbf{h}) - f(\mathbf{0}) - Df(\mathbf{0})\mathbf{h}}{|\mathbf{h}|} =0$
So I did  $\lim\limits_{\mathbf{h} \to 0} \dfrac{f(\mathbf{h}) - Df(\mathbf{0})\mathbf{h}}{|\mathbf{h}|} =0$
Here I am completely stuck, I have no idea how to evaluate the matrix $Df(\mathbf{0})$, so I can't handle the limit. The book uses a different approach, it argues that $Df(\mathbf{0})$ is $0$ (which makes the algebra a lot easier) and they used the alternative limit quotient where no $\mathbf{h}$ appears, but such a quotient is never even mentioned in my book
EDIT1. At some point the book utilizes $|xy| \leq \frac{1}{2}(x^2 +y^2)$. This resembles the AM_GM inequality or is this something else? Or was there some other Lemma that gives rise to this inequality?
EDIT2: I want to mention that I handled the last part of the question already, that is showing it is not $C^1$ in any neighbourhood of $\mathbf{0}$. I just computed $f_x(x,y)$ from first principle and concluded the limit does not even exist. The book chose to look at the limit of $f_x(x,y)$ an interval of $0$ - particular $(0,y)$, but I don't think that's necessary
 A: I will assume we work with the Euclidean norm on $\mathbb{R}^2$: $\|(x,y)\|=\sqrt{x^2+y^2}$.
Recall the inequality $2|xy|\leq x^2+y^2$ which holds for every $x,y\in\mathbb{R}$. This follows by developing the obvious inequality $(|x|-|y|)^2\geq 0$. Note that $|xy|\leq x^2+y^2$ would be sufficient to prove our claim below.
So for all $(x,y)\neq (0,0)$, we have
$$
\frac{|f(x,y)|}{\|(x,y)\|}=\frac{|xy|}{\sqrt{x^2+y^2}}\leq\frac{x^2+y^2}{2\sqrt{x^2+y^2}}=\frac{\sqrt{x^2+y^2}}{2}=\frac{\|(x,y)\|}{2}.
$$
Let $L$ denote the null linear map $L(x,y)=0$.
The inequality above shows that
$$
\lim_{(x,y)\rightarrow(0,0)}\frac{|f(x,y)|}{\|(x,y)\|}=\lim_{(x,y)\rightarrow(0,0)}\frac{|f(x,y)-f(0,0)-L(x,y)|}{\|(x,y)\|}=0.
$$
By definition, this proves that $f$ is differentiable at $(0,0)$ with derivative $Df(0)=L=0$.
Note: the strategy here is to find a candidate for $Df(0)$, and then to check it satisfies the definition of differentiability. If you can't see what $Df(0)$ is gonna be, you can't get started.
Since you say you handled the last part, I'll stop here.
