Answer
$\frac{1}{36}$; $(x-\frac{1}{6})^2$
Work Step by Step
To complete the square, add a term that is the square of half the coefficient of $x$ in the middle term. Then factor by trial and error.
$x^2-\frac{1}{3}x$
$(\frac{-\frac{1}{3}}{2})^2 = (-\frac{1}{3}\times\frac{1}{2})^2=(-\frac{1}{6})^2 = \frac{1}{36}$
$x^2-\frac{1}{3}x+\frac{1}{36}$
$(x-\frac{1}{6})^2$
