Do you mean f(t)=1/t-t?
f(t+k)=1/(t+k)-t-k where k is a small increment.
The rate of change is the derivative=(f(t+k)-f(t))/k in the limit as k→0.
f(t)=(1-t2)/t, f(t+k)=(1-(t+k)2)/(t+k).
Δf=f(t+k)-f(t)=(1-(t+k)2)/(t+k)-(1-t2)/t,
Δf=[t(1-(t+k)2)-(t+k)(1-t2)]/[t(t+k)],
Δf=[t-t3-2kt2-tk2-t-k+t3+kt2]/(t2+kt),
Δf=(-kt2-tk2-k)/(t2+kt).
As k→0, tk2→0 and kt is very small compared to t2, so:
Δf→(-kt2-k)/t2=-k-k/t2, and Δf/k→df/dt=-1-1/t2.