prove that $f(x,y)=|xy|$ is differentiable at (0,0) using $\epsilon_1,\epsilon_2$ definition [closed]

I think we can just take $$ \epsilon_1 = \frac{|\Delta x \Delta y|}{2\Delta x}\\ \epsilon_2 = \frac{|\Delta x \Delta y|}{2\Delta y} $$ $\newcommand{\sgn}{\textrm{sgn}}$ Recalling that $|a| = \sgn(a)a$, for all nonzero $\Delta x $ and $ \Delta y$, $$ \epsilon_1 = \frac{1}{2}\sgn(\Delta x)|\Delta y| $$ and $|\epsilon_1| = |\Delta y|/2$. Therefore, $$ \lim_{\Delta x \to 0}\left(\lim_{\Delta y \to 0} \epsilon_1\right) = 0 $$ The process similarly applies to $\epsilon_2$.

I would recommend computing partial derivatives and then checking that the limit definition you have works out. In particular, $$f_x(0,0) = \lim_{\Delta x\to 0}\frac{f(0+\Delta x,0)-f(0,0)}{\Delta x} = \lim_{\Delta x\to 0} \frac 0{\Delta x} = 0,$$ and similarly for $f_y$. So you need to look at $$\Delta z = |\Delta x\Delta y| - 0\Delta x - 0\Delta y = |\Delta x\Delta y|.$$ Now this can be split up in zillions of different ways as $\epsilon_1\Delta x+\epsilon_2\Delta y$. For example, take $\epsilon_1 = \frac12\eta_1|\Delta y|$ and $\epsilon_2 = \frac12\eta_2|\Delta x|$, where $\eta_1 = \dfrac{|\Delta x|}{\Delta x}$ whenever $\Delta x\ne 0$ (and say $0$ when $\Delta x = 0$), and similarly for $\eta_2$. This is actually quite an awkward definition to work with in this case because of the absolute values.

For your reference, a more common definition, which is easier to work with, is that we should have $$\lim_{(\Delta x,\Delta y)\to (0,0)} \frac{\Delta z - f_x(x_0,y_0)\Delta x - f_y(x_0,y_0)\Delta y}{\sqrt{(\Delta x)^2 + (\Delta y)^2}} = 0.$$ In our case, we only have to check that $$\lim_{(\Delta x,\Delta y)\to (0,0)} \frac{|\Delta x\Delta y|}{\sqrt{(\Delta x)^2 + (\Delta y)^2}} = 0.$$ Can you do this?