First find a common denominator and combine the fractions in the numerator:
\(\displaystyle\lim_{x\to0}\frac{\dfrac1{3+x}-\dfrac13}x=\lim_{x\to0}\frac{\dfrac3{3(3+x)}-\dfrac{3+x}{3(3+x)}}x=\lim_{x\to0}\frac{3-(3+x)}{3x(3+x)}\)
Now simplify and cancel out all the terms that you can:
\(\displaystyle\lim_{x\to0}\frac{3-3-x}{3x(3+x)}=-\frac13\lim_{x\to0}\frac1{3+x}\)
Since the remaining expression is continuous as a function of \(x\), you can directly substitute to end up with
\(\displaystyle-\frac13\lim_{x\to0}\frac1{3+x}=-\frac13\times\frac1{3+0}=-\frac19\)
