Given $f:R_* \rightarrow R_*$ defined by $f(x)=\Large \frac{1}{x}$ where $R_*$ is a set of nonzero real numbers:

Let $x$ and $y$ be two elements in $R_*$. For a one-one function, $f(x) = f(y)$

$ \Rightarrow \Large \frac{1}{x}=\frac{1}{y}$$ \Rightarrow x = y.$

Therefore $f:R_* \rightarrow R_*$ defined by $f(x)=\Large \frac{1}{x}$ is one-one.

For an on-to function, for every $y \in Y$, there exists an element x in X such that $f(x) = y$.

$ \Rightarrow$ For every $y \in R_*$ there must exist $ x=\large \frac{1}{y}$$ \in R_*$ such that $ f(x)=\frac{1}{\Large ( \frac{1}{y})}$$=y$.

Therefore $f:R_* \rightarrow R_*$ defined by $f(x)=\Large \frac{1}{x}$ is onto.

Solution: $f:R_* \rightarrow R_*$ defined by $f(x)=\Large \frac{1}{x}$ is one-one and onto.