Prove that a function $g$ has no roots

As stated, there is no answer to the proposed claim because the domain of any such $g$ must be a subset of $\Bbb R-\{0\}$ for it to be defined. Thus if $g$ is defined, the range of the function $f$ must be a subset of $\Bbb R - \{0\}$.

To prove that $g$ has no roots, we can proceed as follows.

If we write $y = f(x)$, then $g(x) = 1/f(x) = 1/y$. Now we claim that $1/y \ne 0$. If this were so, then we could multiply by $y^2$ on both sides of our equation to obtain $0 = y^2\cdot 0 = y^2\cdot 1/y = y$. However, this says that $y = 0$, which is impossible because $y$ belongs to the range of $f$. We conclude that $g$ can have no roots.

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Also, you apparently have some misunderstanding of the definition of a function. For instance, the map $x\mapsto 1/x^2$ is not a function $\Bbb R\to \Bbb R$, since it is not defined for $x = 0$.