0 is correct, though...
[tex]\dfrac{\frac1x - \frac13}{x + 3}[/tex]
is continuous at [tex]x=3[/tex], so
[tex]\displaystyle \lim_{x\to3} \frac{\frac1x - \frac13}{x + 3} = \frac{\frac13 - \frac13}{\frac13 + 3} = \frac0{\frac{10}3} = 0[/tex]
What was probably intended was the limit
[tex]\displaystyle \lim_{x\to3} \frac{\frac1x - \frac13}{x - 3}[/tex]
which requires a little more finesse because the function is not continuous at [tex]x=3[/tex].
Rewrite it as
[tex]\dfrac{\frac1x - \frac13}{x - 3} = \dfrac{3 - x}{3x(x-3)} = -\dfrac{x-3}{3x(x-3)}[/tex]
When [tex]x\neq3[/tex], we can cancel [tex]x-3[/tex] so that
[tex]\displaystyle \lim_{x\to3} \frac{\frac1x-\frac13}{x-3} = \lim_{x\to3} -\frac1{3x} = -\frac1{3\times3} = -\frac19[/tex]
