You use the foil method as you would when multiplying polynomials of integer exponents. for a pair of binomials the foil method is (a+b)(c+d) = a(c+d) + b(c+d) = ac + ad + bc + bd (rearranging for like terms afterward)
so in this case we have
$\displaystyle (2x^\frac{1}{2}-3y^\frac{3}{2})(2x^\frac{1}{2} + 3y^\frac{3}{2})$
$\displaystyle (2x^\frac{1}{2})(2x^\frac{1}{2}) + (2x^\frac{1}{2})(3y^\frac{3}{2}) - (2x^\frac{1}{2})(3y^\frac{3}{2}) - (3y^\frac{3}{2})(3y^\frac{3}{2})$
$\displaystyle (2x^\frac{1}{2})(2x^\frac{1}{2}) - (3y^\frac{3}{2})(3y^\frac{3}{2})$
$\displaystyle 4x - 9y^3$ by exponent rules. Recall $\displaystyle (a^b)(a^c) = a^{b+c}$