Answer
$$f'\left( x \right) = \frac{1}{{2\sqrt x }},\,\,\,\,f'\left( { - 2} \right){\text{ does not exist}},\,\,\,\,f'\left( 0 \right){\text{ does not exist}},{\text{ and }}f'\left( 3 \right) = \frac{1}{{2\sqrt 3 }}$$
Work Step by Step
$$\eqalign{
& f\left( x \right) = \sqrt x \cr
& {\text{Using the definition of the derivative}}: \cr
& {\text{Find }}f\left( {x + h} \right) \cr
& f\left( {x + h} \right) = \sqrt {x + h} \cr
& {\text{Find }}f\left( {x + h} \right) - f\left( x \right) \cr
& f\left( {x + h} \right) - f\left( x \right) = \sqrt {x + h} - \sqrt x \cr
& {\text{Divide by }}h{\text{ to get }}\frac{{f\left( {x + h} \right) - f\left( x \right)}}{h} \cr
& \frac{{f\left( {x + h} \right) - f\left( x \right)}}{h} = \frac{{\sqrt {x + h} - \sqrt x }}{h} \cr
& {\text{Rationalizing}} \cr
& \frac{{f\left( {x + h} \right) - f\left( x \right)}}{h} = \frac{{\sqrt {x + h} - \sqrt x }}{h} \times \frac{{\sqrt {x + h} + \sqrt x }}{{\sqrt {x + h} + \sqrt x }} \cr
& \frac{{f\left( {x + h} \right) - f\left( x \right)}}{h} = \frac{{{{\left( {\sqrt {x + h} } \right)}^2} - {{\left( {\sqrt x } \right)}^2}}}{{h\left( {\sqrt {x + h} + \sqrt x } \right)}} \cr
& \frac{{f\left( {x + h} \right) - f\left( x \right)}}{h} = \frac{{x + h - x}}{{h\left( {\sqrt {x + h} + \sqrt x } \right)}} \cr
& \frac{{f\left( {x + h} \right) - f\left( x \right)}}{h} = \frac{1}{{\sqrt {x + h} + \sqrt x }} \cr
& {\text{Let }}h \to 0 \cr
& f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {x + h} \right) - f\left( x \right)}}{h} = \mathop {\lim }\limits_{h \to 0} \frac{1}{{\sqrt {x + h} + \sqrt x }} \cr
& {\text{Then}}{\text{,}} \cr
& f'\left( x \right) = \frac{1}{{\sqrt {x + 0} + \sqrt x }} \cr
& f'\left( x \right) = \frac{1}{{2\sqrt x }} \cr
& \cr
& {\text{Find }}f'\left( { - 2} \right),f'\left( 0 \right){\text{ and }}f'\left( 3 \right) \cr
& f'\left( { - 2} \right) = \frac{1}{{2\sqrt { - 2} }},{\text{ The derivative does not exist}} \cr
& f'\left( 0 \right) = \frac{1}{{2\sqrt 0 }},{\text{ The derivative does not exist}} \cr
& f'\left( 3 \right) = \frac{1}{{2\sqrt 3 }} \cr} $$
