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I know on $\textbf{R}$, a continuous function on a bounded set is not necessarily bounded, e.g. $f(x)=\frac{1}{x}$ on $(0,1)$.

Is this true on $\textbf{R}^n$, i.e.let $A\subset \textbf{R}^n$ is bounded, but $f(A)$ is not necessarily bounded. If so, how to prove it?

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    $1/x$+$1/y$ on $(0,1)^2$? – 2017-02-27

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Obviously we need to assume non-compactness. By Heine-Borel, there exists $x_0 \in \bar{A}\setminus A$. Put $$ f(x) = \frac{1}{\lVert{x - x_0\rVert}}. $$ This is a generalizaton of your example in $\mathbb{R}$. In fact, this shows a partial converse of the "continuous functions attain extrema on compact sets" statement: if every continuous function on a set is bounded, then the set is compact.