MATH question

(1)
x = 1/sqrt(y^2 - 5)
x^2 = 1/(y^2 - 5)
x^2(y^2 - 5 ) = 1
x^2y^2 - 5x^2 = 1
x^2y^2 = 1 + 5x^2
y^2 = (1 + 5x^2)/x^2
y = +/- sqrt(1 + 5x^2)/x
so the inverse function is x = +/- sqrt(1 + 5y^2)/y.
(2)
x(3y + 2) = (4y - 3)^3
(3y + 2) + x(3y') = 3(4y - 3)^2(4y')
(3y + 2) = [12(4y - 3)^2 - 3x]y'
so dy/dx = y' = (3y + 2)/[12(4y - 3)^2 - 3x]
when y = 1,
5x = 1, so x = 1/5.
so dy/dx when y = 1 is 5/(12 - 3/5) = 25/(60 - 3) = 25/57.