Answer
$$\frac{1}{2}\frac{e^2-1}{e}$$
Work Step by Step
Since the function $ \frac{e^{x^2}-e^{-y^2}}{x+y}$ is continuous at $(1,1)$, then we have
$$ \lim\limits_{(x,y) \to (1,1)} \frac{e^{x^2}-e^{-y^2}}{x+y}= \frac{e^{1}-e^{-1}}{1+1}=\frac{1}{2}\frac{e^2-1}{e}.$$
