Finding the derivative of $N$ with respect to $t$ of $N= 500\left(1-\frac{3}{(t^2+2)^2}\right)$
The first step to simplify things is to note that since $\frac{3}{(t^2+2)^2}=3(t^2+2)^{-2}$, we have $$N= 500\left(1-\frac{3}{(t^2+2)^2}\right) = 500(1-3(t^2+2)^{-2}) $$
Now let's take the derivative of $N$ with respect to $t$: \begin{align} \frac{dN}{dt} &= \frac{d}{dt}500(1-3(t^2+2)^{-2}) \\ &= 500\frac{d}{dt}(1-3(t^2+2)^{-2}) \\ &= 500 \left(\frac{d}{dt}(1) - \frac{d}{dt}3(t^2+2)^{-2} \right) \\ &= 500\left(0 - 3\frac{d}{dt}(t^2+2)^{-2}\right) \\ &= 500\left( -3\frac{d}{dt}(t^2+2)^{-2} \right) \\ &= -1500\frac{d}{dt}(t^2+2)^{-2} \end{align}
Nowe have to do the chain rule to evaluate \begin{align} \frac{d}{dt}(t^2+2)^{-2} &= -1500\left( -2(t^2+2)^{-2-1}\frac{d}{dt}(t^2+2) \right) \\ &= 3000(t^2+2)^{-3}(2t+0) \\ &= \frac{6000t}{(t^2+2)^3} \end{align}
Now, if we want to know what the rate of change of $N$ with respect to $t$ is when t is say 2, then we plug $t=2$ into the formula we just derived: $$= \frac{6000(2)}{(2^2+2)^3} = 55.555555 \approx 55.6$$
Your answer is correct, and it is important initially to write out all the steps, just like you have done. Note that the derivative is positive or negative depending upon whether or not $t$ is, so from the derivative (and from the original function as well) it is found that $N$ has a minimum at $0$, where $N = 125$. Furthermore, we conclude that $N$ is increasing or decreasing in a neighbourhood of $t$, if $t$ is positive or negative respectively.