Answer
$$
f(x)=\frac{6}{x}
$$
Now use the rules for limits to get
$$
\begin{split}
f^{\prime}(x) &=\lim\limits_{h \to 0} \frac{f(x +h)- f(x)}{h}\\
&=\lim\limits_{h \to 0} \frac{\frac{6}{x+h}-\frac{6}{x}}{h}\\
&=\lim\limits_{h \to 0} \frac{-6h}{xh(x+h)}\\
&=\lim\limits_{h \to 0} \frac{-6}{x(x+h)} \\
&=\frac{-6}{x^{2}}
\end{split}
$$
Use this result to find
$$
f^{\prime}(2)=\frac{-6}{(2)^{2}}=\frac{-3}{2}
$$
$$
f^{\prime}(16)=\frac{-6}{(16)^{2}}=\frac{-6}{4}=\frac{-6}{256}= \frac{-3}{128}
$$
and
$$
f^{\prime}(-3)=\frac{-6}{(-3)^{2}}= \frac{-6}{9}=\frac{-2}{3}
$$
Work Step by Step
$$
f(x)=\frac{6}{x}
$$
Go through the four steps to find $f^{\prime}(x)$
Step 1 Find $f(x+h)$ by replacing $x$ with $x +h$
$$
f(x +h)=\frac{6}{x+h}
$$
Step 2
$$
f(x +h) -f(x)=\frac{6}{x+h}-\frac{6}{x}
$$
Step 3
$$
\frac{f(x +h)- f(x)}{h}= \frac{\frac{6}{x+h}-\frac{6}{x}}{h}
$$
Step 4 Now use the rules for limits to get
$$
\begin{split}
f^{\prime}(x) &=\lim\limits_{h \to 0} \frac{f(x +h)- f(x)}{h}\\
&=\lim\limits_{h \to 0} \frac{\frac{6}{x+h}-\frac{6}{x}}{h}\\
&=\lim\limits_{h \to 0} \frac{-6h}{xh(x+h)}\\
&=\lim\limits_{h \to 0} \frac{-6}{x(x+h)} \\
&=\frac{-6}{x^{2}}
\end{split}
$$
Use this result to find
$$
f^{\prime}(2)=\frac{-6}{(2)^{2}}=\frac{-3}{2}
$$
$$
f^{\prime}(16)=\frac{-6}{(16)^{2}}=\frac{-6}{4}=\frac{-6}{256}= \frac{-3}{128}
$$
and
$$
f^{\prime}(-3)=\frac{-6}{(-3)^{2}}= \frac{-6}{9}=\frac{-2}{3}
$$
