show that y = (kx^2-1)/(kx^2+1) has exactly one stationary point when k is non-zero.

Stationary points are found by considering the points at which the gradient of the function equal zero. For the above, you need to employ the quotient rule, since both numerator and denominator are f(x), to find dy/dx (=y'). If y = u/v, then y' = (u'v - v'u) /v^2. and so y' = 4kx/(kx^2+1)^2. Stationary points are when y' = 0. The numerator is the only important term here, and 4kx = 0 only when x = 0 for all non-zero constants of k - i.e. one stationary point.