How do I find the derivative of these two problems

y = (x^2 - 9)^7/(2x - 3)^3


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xy + (x+1)/(y-1) = 12

Thanks

y = (x^2 - 9)^7/(2x - 3)^3

Apply the quotient rule:

dy/dx = [ (2x-3)^3 (7) (x^2-9)^6 (2x) - (x^2-9)^7 (3) (2x-3)^2 (2)] / (2x-3)^6

The numerator can be factored as: (2x-3)^2 (x^2-9)^6 [ 14x(2x-3) - 6(x^2-9)]
= (2x-3)^2 (x^2-9)^6 (28x^2-42x-6x^2+54)
= (2x-3)^2(x^2-9)^6 (22x^2-42x+54)

dy/dx = (2x-3)^2(x^2-9)^6 (22x^2-42x+54) / (2x-3)^6
dy/dx = (x^2-9)^6 (22x^2-42x+54) / (2x-3)^4
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xy + (x+1) /(y-1) = 12

xy + (x+1) (y-1)^(-1) = 12
differentiate both sides with respect to x

y + x y' + (1) (y-1)^(-1) + (x+1) (-1) (y-1)^(-2) y' = 0
y + x y' - (y-1)^(-1) - (x+1)(y-1)^(-2) y' = 0

y' (x - (x+1)(y-1)^(-2) ) = -y + (y-1)^(-1)

y' = [(y-1)^(-1) - y ] / [ x - (x+1)(y-1)^(-2)]
y' = [ 1/(y-1) - y ] / [ x - (x+1)/(y-1)^2]