Answer
Continuous at $R^{2}$
Work Step by Step
Given: $f(x,y)=\frac{xy}{1+e^{x-y}}$
The given function is defined for all values of x and y except at $1+e^{x-y}=0$
This implies $e^{x-y}=-1$
But the value of $e^{x-y}$ cannot be negative; it is always greater than zero. This implies that $e^{x-y}>0$.
Hence, the function $f(x,y)=\frac{xy}{1+e^{x-y}}$ is continuous at $R^{2}$.
