Verified answer

The answer to this problem is : The bottom by the derivative of the top, minus the top by the derivative of the bottom, all over the bottom to be squared.

I have no paper near me, but here we go.

The derivative of the top is simply 2x

The derivative of the bottom is 2(x-3)(1) = 2(x-3) = 2x-6

So we have: (x-3)²(2x) - (x²-6x)(2x-6) / ((x-3)²)²

I hope you can solve it from there.

Use quotient rule:

dy/dx = [(x-3)^2 * (2x-6) - (x^2-6) * 2(x-3)]/((x-3)^4

I'll leave it up to you to simplify this.

2x - 6(x - 3)² - 2(x² -6x)(x - 3)/(x - 3)^4

2x - 6x² +36x - 5 - 2x^3 + 6x² + 12x² - 36x / (x - 3)^4

-2x^3 + 12x² + 2x - 5 / (x - 3)^4

[x^2-6x] / x^2-6x+9

= [[x^2-6x+9]* d/dx[x^2-6x] - [x^2-6x]*d/dx[x^2-6x+9]] / [x^2-6x+9]^2

= [x^2-6x+9][2x-6] - [x^2-6x][2x-6]

=-72x+18x-54

= -54x-54 / [x^2-6x+9]^2