We have to find 2nd derivative,
f'(x) = (x^2-5 -x*2x)/(x^2-5)^2 = (-x^2-5)/(x^2-5)^2
f''(x) = (-2x(x^2-5)^2-(-x^2-5)2(x^2-5)*2x)/(x^2-5)^4
f''(x) = (-2x(x^2-5)-(-x^2-5)2*2x)/(x^2-5)^3
f''(x) = (-2x^3+10x+4x^3+20x)/(x^2-5)^3
f''(x) = (2x^3+30x)/(x^2-5)^3 = (2x(x^2+15))/(x^2-5)^3
f''(x)=0 <=> 2x=0 (x^2+15 !=0 AAx in R)
f''(x)=0 <=> x=0
(x^2-5)^3 > 0 AAx in (-oo, -sqrt5) uu (sqrt5, +oo)
(x^2-5)^3 < 0 AAx in (-sqrt5,sqrt5)
f''(x)>0 AAx in (-sqrt5,0) uu (sqrt5, +oo) function is concave up
f''(x)<0 AAx in (-oo,-sqrt5) uu (0, sqrt5) function is concave down