We know that #f# has a local maxima /minima at #x=a#, then,
#(1) f'(a)=0, and,
(2) f''(a) lt or gt 0" according as maxima or minima, resp."#
Now, #f(x)=x/(x^2+81) :. f'(x)=((x^2+81)(1)-x(2x))/(x^2+81)^2, i.e.,#
#f'(x)=(81-x^2)/(x^2+81)^2#.
#:. f'(x)=0 rArr x=+-9#
Further, #f''(x)={(x^2+81)^2(-2x)-(81-x^2)2(x^2+81)2x}/(x^2+81)^4#
#={(x^2+81)(-2x)(x^2+81+2(81-x^2))}/(x^2+81)^4#
#={-2x(3(81)-x^2)}/(x^2+81)^3#
Now, #f''(9)={-2(9)(3(81)-81)}/(81+81)^3={-2(9)(2(81))}/(2^3*81^3) lt 0#
#rArr f(9)=9/(81+81)=9/(2*81)=1/18" is local Maxima."#
Finally, #f''(-9)={-2(-9)(2(81))}/(2^3*81^3) gt 0#
#rArr f(-9)=-9/(2*81)=-1/18" is local Minima."#