Aligning two sets of equations with alignat?

You have too many &'s. Just three per row suffice.

On the other hand, aligning at the = signs doesn't seem the best idea, as it creates irregular shapes and the = signs are not really related to one another, so I propose a different solution.

I also added \, after the ! when the factorial is immediately followed by another term.

\documentclass{article}
\usepackage{mathtools}

\begin{document}

\begin{alignat*}{2}
f(x)&=x^{-1},         \qquad & f(2) &= 2^{-1}=\frac{1}{2},
\\
f'(x)&=-x^{-2},       \qquad & f'(2) &=-\frac{1}{2^{2}},
\\
f''(x)&=2!\,x^{-3},   \qquad & \frac{f''(2)}{2!}&=2^{-3}=\frac{1}{2^{3}},
\\
f'''(x)&=-3!\,x^{-4}, \qquad & \frac{f'''(2)}{3!}&=-\frac{1}{2^{4}},
\\
&\vdotswithin{=}      \qquad & & \vdotswithin{=}
\\
f^{(n)}(x)&=(-1)^{n}n!\,x^{-(n+1)}, \qquad &  \frac{f^{(n)}(2)}{n!}&=\frac{(-1)^{n}}{2^{n+1}}.
\end{alignat*}

\begin{alignat*}{2}
& f(x)=x^{-1},          && f(2) = 2^{-1}=\frac{1}{2},
\\[1ex]
& f'(x)=-x^{-2},        && f'(2) =-\frac{1}{2^{2}},
\\[1ex]
& f''(x)=2!\,x^{-3},    && \frac{f''(2)}{2!}=2^{-3}=\frac{1}{2^{3}},
\\[1ex]
& f'''(x)=-3!\,x^{-4},  && \frac{f'''(2)}{3!}=-\frac{1}{2^{4}},
\\
&\qquad\vdots           && \qquad\vdots
\\
& f^{(n)}(x)=(-1)^{n}n!\,x^{-(n+1)},\qquad  &&  \frac{f^{(n)}(2)}{n!}=\frac{(-1)^{n}}{2^{n+1}}.
\end{alignat*}

\end{document}

OK, here we go. A simple align (of course with the appropriately set &) yields

\documentclass{article}
\usepackage{mathtools}
\begin{document}
\begin{align}
        f(x)&=x^{-1}, & f(2) &= 2^{-1}=\frac{1}{2},\\
        f'(x)&=-x^{-2},  &f'(2) &=-\frac{1}{2^{2}},\\
        f''(x)&=2!x^{-3},  &\frac{f''(2)}{2!}&=2^{-3}=\frac{1}{2^{3}},\\
        f'''(x)&=-3!x^{-4},  &\frac{f'''(2)}{3!}&=-\frac{1}{2^{4}},\\
        &\vdotswithin{=} & &\vdotswithin{=}\\
        f^{(n)}(x)&=(-1)^{n}n!x^{-(n+1)}, &  \frac{f^{(n)}(2)}{n!}&=\frac{(-1)^{n}}{2^{n+1}}
\end{align}
\end{document}