Answer
$$
f(x)=-3\sqrt {x}
$$
Now use the rules for limits to get
$$
\begin{split}
f^{\prime}(x) &=\lim\limits_{h \to 0} \frac{f(x +h)- f(x)}{h}\\
&= \lim\limits_{h \to 0}\frac{-3\sqrt {x+h} +3\sqrt {x}}{h}\\
&=\lim\limits_{h \to 0}\frac{-3\sqrt{x+h}+3\sqrt{x}}{h} \cdot \frac{-3\sqrt{x+h}-3\sqrt{x}}{-3\sqrt{x+h}-3\sqrt{x}} \\
&=\lim\limits_{h \to 0} \frac{9(x+h)-9x}{h(-3\sqrt{x+h}-3\sqrt{x})} \\
&=\lim\limits_{h \to 0} \frac{9h}{h(-3\sqrt{x+h}-3\sqrt{x}) } \\
&=\lim\limits_{h \to 0} \frac{9}{-3\sqrt{x+h}-3\sqrt{x}}\\
&=-\frac{3}{2\sqrt {x}}
\end{split}
$$
Use this result to find
$$
f^{\prime}(2)=-\frac{3}{2\sqrt {2}}
$$
$$
f^{\prime}(16)=-\frac{3}{2\sqrt {16}}=\frac{-3}{8}
$$
and
$$
f^{\prime}(-3)=-\frac{3}{2\sqrt {-3}} \\
\quad\quad\quad\quad[\text { is not a real number , hence }]
$$
$f^{\prime}(-3)$ does not exist.
Work Step by Step
$$
f(x)=-3\sqrt {x}
$$
Go through the four steps to find $f^{\prime}(x)$
Step 1 Find $f(x+h)$ by replacing $x$ with $x +h$
$$
f(x +h)=-3\sqrt {x+h}
$$
Step 2
$$
f(x +h) -f(x)=-3\sqrt {x+h} -(-3\sqrt {x})
$$
Step 3
$$
\frac{f(x +h)- f(x)}{h}= \frac{-3\sqrt {x+h} +3\sqrt {x}}{h}
$$
Step 4 Now use the rules for limits to get
$$
\begin{split}
f^{\prime}(x) &=\lim\limits_{h \to 0} \frac{f(x +h)- f(x)}{h}\\
&= \lim\limits_{h \to 0}\frac{-3\sqrt {x+h} +3\sqrt {x}}{h}\\
&=\lim\limits_{h \to 0}\frac{-3\sqrt{x+h}+3\sqrt{x}}{h} \cdot \frac{-3\sqrt{x+h}-3\sqrt{x}}{-3\sqrt{x+h}-3\sqrt{x}} \\
&=\lim\limits_{h \to 0} \frac{9(x+h)-9x}{h(-3\sqrt{x+h}-3\sqrt{x})} \\
&=\lim\limits_{h \to 0} \frac{9h}{h(-3\sqrt{x+h}-3\sqrt{x}) } \\
&=\lim\limits_{h \to 0} \frac{9}{-3\sqrt{x+h}-3\sqrt{x}}\\
&=-\frac{3}{2\sqrt {x}}
\end{split}
$$
Use this result to find
$$
f^{\prime}(2)=-\frac{3}{2\sqrt {2}}
$$
$$
f^{\prime}(16)=-\frac{3}{2\sqrt {16}}=\frac{-3}{8}
$$
and
$$
f^{\prime}(-3)=-\frac{3}{2\sqrt {-3}} \\
\quad\quad\quad\quad[\text { is not a real number , hence }]
$$
$f^{\prime}(-3)$ does not exist.