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."