How do I differentiate this probleM

y= (x^2-6x-9)/sqrt(x)

how do I differentiate this?
I used the quotient rule and got (3(x^2-6x-3))/(2xsqrt(x)), which turns out to be incorrect. Can anyone help me out pls? Thanks :)

Here is another method. Just separate out the fraction
y= (x²-6x-9) /√x

= x²/√x - 6x/√x - 9/√x
= x^(3/2) - 6x^(1/2) -9x^(-1/2), so

dy/dx = (3/2)x^(1/2) -3x^(-1/2) +(9/2)x^(-3/2)

Given : y= (x^2-6x-9)/√x
dy/dx=d/dx
((x^2-6x-9)/√x)

Use the product rule
d/dx(uv)=vdu/dx+udv/dx

Where u=1/√x
v=x²-6x-9

There for
y'=3(x²-2x+3)/(2x)^(3/2)

If you had used the Quotient Rule correctly you would have obtained

[√x(2x - 6) - (x² - 6x - 9)(1/2)x^(-1/2)] / x

which could be tidied up.