derivative

Using power rule and reciprocal rule. f '(t) = 2(1/(t-3))[(1/t-3)^2] -2/(t-3)^3

$f(t) = (\frac {1}{t-3})^2 = \frac {1}{(t-3)^2}$

$g(t) = \frac {1}{t^2}$ ; $g'(t)= -2 \times \frac {1}{t^3}$

$h(t) = t-3$ ; $h'(t) = 1$

$f(t) = g(h(t))$

Applying chain rule:

$f'(t) = g'(h(t)) \times h'(t) = (-2 \times \frac{1}{(t-3)^2}) \times 1 = (-2 \times \frac{1}{(t-3)^2})$