Answer
The composition $f\circ g$ is discontinuous on the circle $$x^2+y^2=1.$$
Work Step by Step
The function $g(x,y)=x^2+y^2$ is continuous as a sum of continuous functions $x^2$ and $y^2$. The function $f(t)=\frac{1}{1-t}$ is discontinuous when its' numerator is zero i.e. when $t=1$ because, since we cannot divide by zero, we have
$$\lim_{t\to1}f(t)\neq f(1).$$
The composition $$f\circ g(x,y)=f(g(x,y))$$ is thus discontinuous when the argument of $f$, that is $g(x,y)$, is zero which is for $$g(x,y)=x^2+y^2-1=0\rightarrow x^2+y^2=1$$
The equation $x^2+y^2=1$ represents a circle in $xy$ plane centered at the origin with the radius of $1$ and along this circle the composition is discontinuous.
