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