Find the equation of the normal to the curve $y = 8/(4 + x^2)$ , at $x = 1$. When you first differentiate the above, you get $-8/25$, right? Then you derive the gradient for a normal and proceed so on and so forth.
The textbook I'm using says when you differentiate, you get $-16/25$. I believe that's wrong...

The textbook is correct. 
You can use
$$\left\{\frac{f(x)}{g(x)}\right\}^\prime=\frac{f^\prime(x)g(x)-f(x)g^\prime(x)}{\{g(x)\}^2}.$$
Letting $$h(x)=\frac{8}{4+x^2},$$
we have 
$$h^\prime(x)=\frac{0-8\cdot 2x}{(4+x^2)^2}=-\frac{16x}{(4+x^2)^2}.$$
Hence, we will have 
$$h(1)=-\frac{16}{25}.$$
So...
