Answer
The limit is
$$\lim_{(x,y)\to(0,1)}\frac{\arcsin xy}{1-xy}=0.$$
The function is discontinuous on a hyperbola $y=\frac{1}{x}$.
Work Step by Step
We will calculate this limit by substitution:
$$\lim_{(x,y)\to(0,1)}\frac{\arcsin xy}{1-xy}=\frac{\arcsin(0\cdot1)}{1-0\cdot1}=\frac{\arcsin 0}{1}=0.$$
The function will have a discontinuity when $1-xy=0$ which is when $xy=1\Rightarrow y=1/x$. This is geometrically a hyperbola in the $xy$ plane. The discontinuity happens because when $1-xy=0$ the denominator iz $0$ and since we cannot divide by zero will have
$$\lim_{(x,y)\to(x_0,y_0)}\frac{\arcsin xy}{1-xy}\neq\frac{\arcsin x_0y_0}{1-x_0y_0}.$$
At all other points from the domain we can just substitute and get
$$\lim_{(x,y)\to(x_0,y_0)}\frac{\arcsin xy}{1-xy}=\frac{\arcsin x_0y_0}{1-x_0y_0},$$
so there the function would be continuous.