Answer
The limit is
$$\lim_{(x,y)\to(-1,2)}\frac{x+y}{x-y}=-\frac{1}{3}.$$
The function has a discontinuity on the plane $x=y$.
Work Step by Step
The limit can be calculated by simple substitution
$$\lim_{(x,y)\to(-1,2)}\frac{x+y}{x-y}=\frac{-1+2}{-1-2}=-\frac{1}{3}.$$
The function is not continuous on the set of ordered pairs $(x_0,y_0)$ such that $x=y$ because then we would have zero in the denominator and we would have
$$\lim_{(x,y)\to(x_0,y_0)}\frac{x+y}{x-y}\neq\frac{x_0+y_0}{x_0-y_0},$$
because we cannot divide by zero. In all other points $(x_0,y_0)$, $x_0\neq y_0$ the function is continuous because
$$\lim_{(x,y)\to(x_0,y_0)}\frac{x+y}{x-y}=\frac{x_0+y_0}{x_0-y_0}.$$
