$f(x,y)=2e^{-xy}$
$\frac{\partial}{\partial x}f=-2e^{-xy}*y$
$\frac{\partial}{\partial y}f=-2e^{-xy}*x$
These partial derivatives are continuous, so function is differentiable on $\mathbb R^2$
Consider this method:
Suppose I want to check the differentiability at point $T=(0,1)$ using this method:
$f(0+\Delta x,1+\Delta y)-f(0,1)=\frac{\partial}{\partial y}f(0,1)*\Delta x+\frac{\partial}{\partial x}f(0,1)*\Delta y+o(\Delta x,\Delta y)$
$\frac{\partial}{\partial x}f(0,1)=-2$
$\frac{\partial}{\partial y}f=0$
$f(0+\Delta x,1+\Delta y)-f(0,1)=2e^{-\Delta x(1+\Delta y)}-2$
$2e^{-\Delta x(1+\Delta y)}-2+2\Delta x=o(\Delta x,\Delta y)$
We now need to show:
$$\lim_{(\Delta x,\Delta y)\rightarrow (0,0)}\frac{2e^{-\Delta x(1+\Delta y)}-2+2\Delta x}{\sqrt{\Delta x^2+\Delta y^2}}=0$$ I can't solve this limit. Wolfram alpha returns "limit doesn't exist". Why can't we prove the differentiability this way even though the function is differentiable?