Answer
The value is
$$\left.\frac{dy}{dt}\right|_{t=1}=-\frac{1}{4}.$$
Work Step by Step
Using the definition of the derivative and putting $t=1$ we have
$$\left.\frac{dy}{dt}\right|_{t=1}=\lim_{h\to0}\frac{f(1+h)-f(1)}{h}=\lim_{h\to0}\frac{\frac{1}{(1+h)+1}-\frac{1}{1+1}}{h}=\lim_{h\to0}\frac{\frac{1}{2+h}-\frac{1}{2}}{h}=\lim_{h\to0}\frac{\frac{2-(2+h)}{2(2+h)}}{h}=\lim_{h\to0}\frac{-h}{2h(2+h)}=\lim_{h\to0}\frac{-1}{2(2+h)}=\frac{-1}{2(2+0)}=\frac{-1}{4}.$$
