By the definition of the derivative,
[tex]f'(x)=\displaystyle\lim_{h\to0}\frac{f(x+h)-f(x)}h[/tex]
[tex]\implies y'=\displaystyle\lim_{h\to0}\frac{(4(x+h)^2-9(x+h)-6)-(4x^2-9x-6)}h[/tex]
Simplify the numerator:
[tex]4(x+h)^2-9(x+h)-6=4(x^2+2xh+h^2)-9(x+h)-6[/tex]
[tex]=4x^2+8xh+4h^2-9x-9h-6[/tex]
Subtracting [tex]4x^2-9x-6[/tex] removes all the terms here not involving [tex]h[/tex], so the limit reduces to
[tex]y'=\displaystyle\lim_{h\to0}\frac{8xh+4h^2-9h}h[/tex]
Cancel the factor [tex]h[/tex] in the numerator and denominator:
[tex]y'=\displaystyle\lim_{h\to0}(8x+4h-9)[/tex]
As [tex]h\to0[/tex], we're left with
[tex]\boxed{y'=8x-9}[/tex]