To determine concavity of a function, you must find the second derivative #f''(x)#, determine its critical points (places where #f''(x)=0# or is not defined), and then examine the sign of #f''(x)# in all intervals defined by those critical points.

#f(x) = x^4 - x^2#

#f'(x) = 4x^3 - 2x#

#f''(x) = 12x^2- 2#

From examination #f''(x)# is a quadratic function, and therefore there are no places where it is undefined. Thus, our only critical points will come from the zeroes of #f''(x)#:

#12x^2-2 = 0#

#12x^2 = 2#

#x^2 = 1/6 :. x=+-1/sqrt(6)~~+-0.408#

Having two zeroes means we have 3 intervals to examine:

#(-oo,-1/sqrt(6)), (-1/sqrt(6),1/sqrt(6)), (1/sqrt(6),oo)#

Choose a single #x# value inside of each interval and evaluate #f''(x)# at that value. If the result is positive, the function #f(x)# is concave up in that interval; if the result is negative, the function is concave down. For simplicity, choose "easy" values of #x# to evaluate:

#f''(-1) = 12(-1)^2-2 = 12-2 = 10 > 0 :. "concave up"#

#f''(0) = 12(0)^2-2 = 0-2 = -2 < 0 :. "concave down"#

#f''(1) = 12(1)^2-2 = 12-2 = 10 > 0 :. "concave up"#

graph{x^4-x^2 [-1.923, 1.923, -0.963, 0.959]}