function differentiable at only one point

Let $f\\colon\\mathbbmss{R}\\to\\mathbbmss{R}$ be the function

f(x)=\\begin{cases}x,&\\mbox{when $x$ is rational},\\\\-x,&\\mbox{when $x$ is irrational}.\\end{cases}

See this entry (http://planetmath.org/FunctionContinuousAtOnlyOnePoint).Let $g\\colon\\mathbbmss{R}\\to\\mathbbmss{R}$ be the function

g(x)=f(x)x.

Then $g$ differentiable at $0$ ,but everywhere else non-differentiable.

Indeed, since

$\\displaystyle g^{\\prime}(0)$	$\\displaystyle=$	$\\displaystyle\\lim_{h\\to 0}\\frac{f(h)h-f(0)0}{h}$
	$\\displaystyle=$	$\\displaystyle\\lim_{h\\to 0}f(h)$
	$\\displaystyle=$	$\\displaystyle 0$

$g$ is differentiable at $0$ .If $g$ would be continuous at $x\eq 0$ , then $f(x)=g(x)/x$ would be continuous at $x$ .This result (http://planetmath.org/DifferentiableFunctionsAreContinuous)implies that $g$ is non-differentiable away from the origin.