Answer
The limit is
$$\lim_{(x,y)\to(1,1)}\frac{xy}{x^2+y^2}=\frac{1}{2}.$$
The function has a discontinuity at the origin $(0,0)$.
Work Step by Step
The limit can be found by a simple substitution
$$\lim_{(x,y)\to(1,1)}\frac{xy}{x^2+y^2}=\frac{1\cdot1}{1^2+1^2}=\frac{1}{2}.$$
The function is discontinuous at $(x,y)=(0,0)$ i.e. at the origin and this happens because we would have zero in the denominator and since we cannot divide by zero we would have
$$\lim_{(x,y)\to(x_0,y_0)}\frac{xy}{x^2+y^2}\neq\frac{x_0y_0}{x_0^2+y_0^2}.$$
For all other points $(x_0,y_0)$ where $x_0$ and $y_0$ are not zero at the same time we would have
$$\lim_{(x,y)\to(x_0,y_0)}\frac{xy}{x^2+y^2}=\frac{x_0y_0}{x_0^2+y_0^2},$$
so the function is continuous in other points.
