Let be $f:\mathbb{R}^2\to\mathbb{R}$ with $f(x,y):=|xy|$.
I guess it is quite known that $f$ ist differentiable at $(0,0)$ but not along $(a,0)$ and $(0,b)$ where $a\neq 0$ and $b\neq 0$. I was wondering if I can show it this way:
$(a,0)$:
Differentiability means that I must find some matrix $A$ such that $$ \lim\limits_{{x\choose y}\to {a\choose 0}}\frac{|xy|-0-A\left({x\choose y}-{a\choose 0}\right)}{\Vert {x\choose y}-{a\choose 0}\Vert_\infty}=0. $$ Now, we plug in $x_n:=a+\frac{1}{n}$ and $y_n:=\frac{1}{n}$ and get \begin{align*} \lim\limits_{n\to\infty}\frac{(\frac{1}{n}+\frac{1}{n^2})-A{\frac{1}{n}\choose \frac{1}{n}}}{\frac{1}{n}}=\lim\limits_{n\to\infty}(1+\frac{1}{n})-A{1\choose 1} \end{align*} If we define $A:={a_1\choose a_2}$ then this yields $$ \lim\limits_{n\to\infty}(1+\frac{1}{n}-a_1-a_2)\overset{???}{=}0 $$ However, we see that this condition can be satisfied by infinitely many values for $a_1,a_2$. This contradicts the uniqueness of the derivative. So $f$ can't be differentiable at those points.
The same argument applies if we look at $(0,b)$.
At $(0,0)$ we get: \begin{align*} \lim\limits_{{x\choose y}\to {0\choose 0}}\frac{|xy|-0-A\left({x\choose y}-{0\choose 0}\right)}{\Vert {x\choose y}-{0\choose 0}\Vert_\infty}=\lim\limits_{{x\choose y}\to {0\choose 0}}\frac{|xy|-{a_1\choose a_2}{x\choose y}}{\Vert {x\choose y}\Vert_\infty}\leq \lim\limits_{{x\choose y}\to {0\choose 0}}|x|+|a_1|+|a_2|. \end{align*} If we choose $a_1=a_2=0$ then the upper bound gets arbitrarily small and delivers the unique derivative $A={0\choose 0}$.
Is this a legit way to prove the differentiability/ non-differentiability at the respective points?
As I mentioned in the comments, you cannot use the fact that your method gives an equation satisfied by many matrices $A$ to conclude that $A$ doesn't exist. All you can conclude is that your method cannot show that $A$ is unique, which doesn't contradict anything.
To put a point on it, you could use that same exact argument on any differentiable $f : \mathbb{R}^2 \to \mathbb{R}$ to fallaciously conclude that it's nowhere differentiable.
Instead, consider the line $x=a$ for $a\neq 0$. If we assume that $f(x) = |xy|$ is differentiable at $(a,0)$, then we conclude that there is some $A = (A_1\; A_2)$ such that $$0 = \lim_{(x,y) \to (a,0)} \frac{|xy| - 0 - A\binom{x-a}{y}}{\|\binom{x-a}{y}\|_\infty} = \lim_{y\to 0} \frac{|ay| - A_2y}{|y|} = \lim_{y\to 0} \left(|a| - A_2\frac{y}{|y|}\right)$$ However, if $y \to 0^+$, then this gives $|a| - A_2 = 0$ and if $y \to 0^-$, this gives $|a| + A_2 = 0$. Adding these gives $2|a| = 0$, which implies $a=0$, which is a contradiction.
Your solution showing $f$ is differentiable at $(0,0)$ has a couple issues, but they're pretty minor in comparison.
First, I'm not sure what your reasoning for $$\frac{|xy|-(a_1\; a_2){x\choose y}}{\Vert {x\choose y}\Vert_\infty}\leq |x|+|a_1|+|a_2|$$ is. It is true, so maybe you just left off that reasoning. From here, I'll assume that is the case.
Second, on a general level, what you've shown is that $F(x,y) \leq G(x,y)$ and $\lim_{(x,y) \to (0,0)} G(x,y) = 0$. From just this, you cannot conclude $\lim_{(x,y) \to (0,0)} F(x,y) = 0$ (you cannot even conclude this limit exists). Instead, what you need to show is $|F(x,y)| \leq G(x,y)$.
Back to your specific question, what you actually want to show is $$\left|\frac{|xy|-(a_1\; a_2){x\choose y}}{\Vert {x\choose y}\Vert_\infty}\right|\leq |x|+|a_1|+|a_2|$$ which is also true, has basically the same proof as the inequality you have, and is sufficient to conclude that $f(x,y)$ is differentiable at $(0,0)$.