I have made an attempt to the following problem. I am not sure about its correctness though. Can you please check if my attempt works:
Let $f:\mathbb R^2\to\mathbb R:(x,y)\mapsto\sqrt{|xy|}.$ Show that $f$ is differentiable at the origin.
My Attempt:
Consider otherwise. If $\lambda:=Df_{(0,0)}$ then $$\lim\limits_{(h,k)\to{(0,0)}}\dfrac{|f(h,k)-f(0,0)-\lambda(h,k)|}{|(h,k)|}=0.$$
Since $\lambda\in\mathcal{L}(\mathbb R^2,\mathbb R),~\exists~A,B\in\mathbb R$ such that $\lambda(h,k)=Ah+Bk~\forall~(h,k)\in\mathbb R^2.$
Note for $m\in\mathbb R,$ $\left(\dfrac{1}{n},\dfrac{m}{n}\right)\to(0,0).$ However
$$\lim\limits_{n\to\infty}\dfrac{\left|f\left(\dfrac{1}{n},\dfrac{m}{n}\right)-f(0,0)-\lambda\left(\dfrac{1}{n},\dfrac{m}{n}\right)\right|}{\left|\left(\dfrac{1}{n},\dfrac{m}{n}\right)\right|}=\lim\limits_{n\to\infty}\dfrac{|\sqrt{m}-A-Bm|}{|\sqrt{1+m^2}|}$$does not exist, a contradiction.
My guess is that you meant to prove that $f$ is not differentiable at $(0,0)$ and that you attemkped to prove it by contradiction.
When you write about $\sqrt m$, this assumes that $m\geqslant0$. You should have stated that. And the assertion that the limit$$\lim\limits_{n\to\infty}\frac{|\sqrt{m}-A-Bm|}{|\sqrt{1+m^2}|}$$does not exist is fals. The limit exists and it is equal to $\frac{|\sqrt{m}-A-Bm|}{|\sqrt{1+m^2}|}$. The problem is that this number depends on $m$, whereas it should not depend on it.