Answer
$(f ◦ g)(x) = \frac{−1}{x}$ with domain $x > 0$, and $(g ◦ f)(x)$ is nowhere defined, with domain $∅$
Work Step by Step
$f(x) = - x^2$, $g(x) = \frac{1}{\sqrt{x}}$
To find $(f ◦ g)(x)$, substitute $g(x)$ into $f(x)$
$(f ◦ g)(x) = - (\frac{1}{\sqrt{x}})^2 = -\frac{1^2}{(\sqrt{x})^2} = -\frac{1}{x}$
To find $(g ◦ f)(x)$, substitute $f(x)$ into $g(x)$
$(g ◦ f)(x) = \frac{1}{\sqrt{-x^2}}$
But we don't have negative under square-root so $(g ◦ f)(x) is not defined
