Answer
The limit is
$$\lim_{(x,y)\to(0,1)}\frac{\arccos(x/y)}{1+xy} =\frac{\pi}{2}.$$
The function is discontinuous along:
1) the line $y=0$ i.e. the $x$ axis;
2) the hyperbola $y=-1/x$.
Work Step by Step
We will calculate this limit by substitution:
$$\lim_{(x,y)\to(0,1)}\frac{\arccos(x/y)}{1+xy} = \frac{\arccos(0/1)}{1+0\cdot1}=\frac{\arccos0}{1}=\frac{\pi}{2}.$$
This function will have two discontinuities:
1) When $y=0$ i.e. the $x$ axis. This is because $y$ appears in the denominator of the argument of $\arccos$ and since we cannot divide by zero we would have
$$\lim_{(x,y)\to(x_0,y_0)}\frac{\arccos(x/y)}{1+xy}\neq \frac{\arccos(x_0/y_0)}{1+x_0y_0},$$
when $y_0=0$.
2) When $1+xy=0$ i.e. when $y=-1/x$ (this is a hyperbola in the $xy$ plane). This is because $1+xy$ is in the denominator, and because we cannot divide by zero we again have
$$\lim_{(x,y)\to(x_0,y_0)}\frac{\arccos(x/y)}{1+xy}\neq \frac{\arccos(x_0/y_0)}{1+x_0y_0},$$
when $$y_0=-1/x_0.$$
At all other points where $y_0\neq0$ and $1+x_0y_0\neq0$ we have that the function is continuous because we can always substitute and get
$$\lim_{(x,y)\to(x_0,y_0)}\frac{\arccos(x/y)}{1+xy}= \frac{\arccos(x_0/y_0)}{1+x_0y_0}.$$
