Answer
$L = \int^{3}_{2} \sqrt{[\frac{1}{2}(x^{2} -2)]^2}dx = \frac{13}{6}$
Work Step by Step
$36y^{2} = (x^2 -4)^{3}, y\geq0$ then $y = 1/6 (x^{2} -4)^{3/2} $ and $dy/dx = \frac{1}{6} \frac{3}{2} (x^{2} -4)^{1/2}(2x) = \frac{1}{2}x(x^{2}-4)^{1/2}$
$1+(dy/dx)^{2} = 1+(1/4)x^2(x^2-4) = (1/4) x^{4} - x^{2} +1 = (1/4)(x^{4} -4x^{2} +4) = [(1/2) (x^{2} - 2)]^2$
Therefore
$L = \int^{3}_{2} \sqrt{[\frac{1}{2}(x^{2} -2)]^2}dx = \frac{1}{2}[(9-6)-(\frac{8}{3}-4)] = \frac{13}{6}$
