Answer
$$ - \frac{1}{9}$$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{x \to 0} \frac{{1/\left( {x + 3} \right) - 1/3}}{x} \cr
& {\text{use the rule }}\mathop {\lim }\limits_{x \to a} \frac{{f\left( x \right)}}{{g\left( x \right)}} = \frac{{\mathop {\lim }\limits_{x \to a} f\left( x \right)}}{{\mathop {\lim }\limits_{x \to a} g\left( x \right)}}\,\,\,\left( {{\text{see page 128}}} \right).{\text{ then}} \cr
& \mathop {\lim }\limits_{x \to 0} \frac{{1/\left( {x + 3} \right) - 1/3}}{x} = \frac{{\mathop {\lim }\limits_{x \to 0} \left( {1/\left( {x + 3} \right) - 1/3} \right)}}{{\mathop {\lim }\limits_{x \to 0} x}} \cr
& {\text{evaluate the limits}}{\text{, replacing }}0{\text{ for }}x \cr
& = \frac{{\left( {1/\left( {0 + 3} \right) - 1/3} \right)}}{0} = \frac{{1/3 - 1/3}}{0} = \frac{0}{0}{\text{ indeterminate form}} \cr
& {\text{Simplifying the numerator as}} \cr
& \frac{1}{{x + 3}} - \frac{1}{3} = \frac{{3 - \left( {x + 3} \right)}}{{3\left( {x + 3} \right)}} = \frac{{ - x}}{{3\left( {x + 3} \right)}} \cr
& then \cr
& \mathop {\lim }\limits_{x \to 0} \frac{{1/\left( {x + 3} \right) - 1/3}}{x} = \mathop {\lim }\limits_{x \to 0} \frac{{\frac{{ - x}}{{3\left( {x + 3} \right)}}}}{x} = \mathop {\lim }\limits_{x \to 0} \frac{{ - x}}{{3x\left( {x + 3} \right)}} \cr
& = \mathop {\lim }\limits_{x \to 0} \frac{{ - 1}}{{3\left( {x + 3} \right)}} \cr
& {\text{evaluating the limit when }}x \to 0 \cr
& = \frac{{ - 1}}{{3\left( {0 + 3} \right)}} \cr
& = - \frac{1}{9} \cr} $$