Solution:

I like to work through these types of questions in a series of steps, beginning with evaluating the function at \(a+h\) and \(a\) and then simplifying the numerator.

\[ \begin{{array}}{ rcl } \displaystyle f({\color{ red } a +h}) &=& ({\color{ red } a+h})^2-1\\ &=&(a+h)(a+h)-1\\ &=&a^2 + ah + ah + h^2 - 1\\ f(a+h)&=&a^2+2ah+h^2 - 1 \end{{array}} \]

The \(f(a)\) evaulation is really just a change of variables, so it is relatively simple to rewrite: \[f({\color{ red } a}) = {\color{ red }a}^2-1\]

Next, I need to figure out \(f(a+h)-f(a)\). Notice that \(f(a)\) has two terms, so when I substitute it, I really need to use extra parenthesis.

\[ \begin{{array}}{{rcl}} {\color{ blue } f(a+h)}-{\color{ green } f(a)} &=& {\color{ blue } a^2+2ah+h^2-1}-({\color{ green } a^2-1})\\ &=&a^2+2ah+h^2-1 - a^2 + 1\\ f(a+h)-f(a) &=&2ah+h^2 \end{{array}} \]

Now that we have evaluated the numerator, let's put it all together and figure out the quotient:

\[\begin{{array}}{{rcl}} \displaystyle \frac{f(a+h)-f(a)}{{h}}&=& \displaystyle\frac{2ah+h^2}{{h}}\\ &=&\displaystyle\frac{h\left(2a+h\right)}{{h}}\\ \displaystyle\frac{f(a+h)-f(a)}{{h}}&=&\boxed{2a+h} \end{{array}} \]