Answer
$$\frac{1}{9}$$
Work Step by Step
We evaluate the limit:
\begin{aligned}
\lim _{x \rightarrow 0}\left(\frac{1}{3 x}-\frac{1}{x(x+3)}\right)&=\lim _{x \rightarrow 0} \frac{x+3-3}{3 x(x+3)} \\
&=\lim _{x \rightarrow 0} \frac{x}{3 x(x+3)}\\
&=\lim _{x \rightarrow 0} \frac{ 1}{3 (x+3)}\\
&=\frac{1}{9}
\end{aligned}
