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