Make "x" the subject of the formula Y=[(b-x²)/(yx²)]⅔
3/2 power ... y^(3/2) = (b-x²) / (yx²)
multiply by y ... y^(5/2) = (b-x²)/(x²) = (b / x^2) - 1
add 1 ... [y^(5/2)] + 1 = b / x^2
multiply by x^2 and divide by [y^(5/2)] + 1 ... x^2 = b / {[y^(5/2)] + 1}
take square root