Answer
The composition $f\circ g$ is discontinuous on the line $y=\frac{2}{3}x.$
Work Step by Step
The function $g(x,y)=2x-3y$ is continuous as a sum of continuous functions $2x$ and $3y$. The function $f(t)=1/t$ has a discontinuity when $t=0$ because we cannot divide by zero and thus $$\lim_{t\to0}f(t)\neq f(0).$$
The composition
$$f\circ g(x,y)=f(g(x,y))$$
is then discontinuous when the argument of $f$, that is $g(x,y)=0$. This is when $2x-3y=0\Rightarrow y=\frac{2}{3}x$ so this function is discontinuous on the line $y=\frac{2}{3}x.$
