How do you find the horizontal and vertical asymptotes of a function with a square root?

When deciding for the asymptotes, you have to look at the values that will make the denominator 0.Now:
sqrt(9x^2-4)=0
9x^2-4=0
9x^2=4
x^2=4/9
x={2/3, -2/3}

The horizontal asymptote is the number that y approaches as x approaches +/- infinity.  For your function, as x gets really large sqrt(9x^2 - 4) gets close to just sqrt(9x^2) = 3x.  The 4 becomes insignificant quickly.  That leaves you with 2x/(3x) which is going to reduce to 2/3.  So the horizontal asymptote will be y = 2/3.

Vertical asyptotes happen when y is +/- infinity.  In order for this to occur, the denominator of your fraction must go to zero.  So what makes sqrt(9x^2 - 4) = 0?  Well, 9x^2 - 4 would have to equal zero.  Thus x = +/- 2/3.

The fact that they are the same is a complete coincidence.