Answer
(a) $x^2$
(b) Yes $f\circ g$ is everywhere continuous
Work Step by Step
\[f(x)=\frac{1}{x}\;\;, \;\; g(x)=\frac{1}{x^2}\]
(a) \[(f\circ g)(x)=f(g(x))\]
\[\Rightarrow (f\circ g)(x)=f(\frac{1}{x^2})\]
\[\Rightarrow (f\circ g)(x)=\frac{1}{\displaystyle\frac{1}{x^2}}=x^2\]
\[(f\circ g)(x)=x^2\]
(b) \[(f\circ g)(x)=x^2\]
$f\circ g$ is polynomial and every polynomial is everywhere continuous,
$\Rightarrow f\circ g$ is everywhere continuous
