Derivates of $f(x,y)=|x| + |y| − ||x| − |y||.$ I worked on one my problem and I want you to check my work or give my some good idea for solution. 
$$f(x,y)=|x| + |y| − ||x| − |y||.$$
a) Find all $(x,y)$ such that $f$ is continuous at given $(x,y)$.
Solution a) it's very easy to see that given function is continuous at every $(x,y)$ because absolute value function $|x|$ is continuous at every $x$.
b) Count partial derivatives of function $f$ at the point $(0,0).$ 
Solution: My idea is not to get partial derivatives for every $x$ and $y$, but just use limit definition of it.
$$\lim_{h\to0} [\frac{f(h,0)-f(0,0)}{h}=\frac{|h|+|0|-||h|-|0||}{h}=\frac{0}{h}=0]$$ This is partial derivatives whith respect of x. Similarly we get $0$ for partial derivatives with respect to y at the point $(0,0)$.
c) Is it function $f$ differentiable at $(0,0)$?
 My work: Idea is to use definition, so we know how looks our differential operator - it's null operator (from b) part).
 $$\lim_{(x,y)\to0} \frac{f(x,y)-f(0,0)-A(x,y)}{||(x,y)||}=\frac{|x| + |y| − ||x| − |y||}{\sqrt{(x^2+y^2)}}$$. If we look at $x=0$ limit is $0$, but if $y=x$ than limit isn't equal to zero, so this limit doesn't exist. 

As you may be aware, if $a$ and $b$ are real numbers, then
\begin{align*}
\max(a, b) &= \tfrac{1}{2}(a + b + |a - b|), \\
\min(a, b) &= \tfrac{1}{2}(a + b - |a - b|).
\end{align*}
(If $I$ denotes the closed interval with endpoint $a$ and $b$, each formula may be interpreted as "starting at the midpoint of $I$ and traveling to the right (max) or left (min) by half the length of $I$.)
Particularly,
$$
f(x, y) = |x| + |y| - \bigl||x| - |y|\bigr| = 2\min(|x|, |y|).
$$
This observation makes your conclusions geometrically apparent.
