Answer
$$\frac{dy}{dx}=-\frac{2(2x+3)^2(52x^2+96x+3)}{(4x^2-1)^9}$$
Work Step by Step
$$\frac{dy}{dx}=\Big(\frac{(2x+3)^3}{(4x^2-1)^8}\Big)'=
\frac{((2x+3)^3)'(4x^2-1)^8-(2x+3)^3((4x^2-1)^8)'}{(4x^2-1)^{16}}=
\frac{3(2x+3)^2(2x+3)'(4x^2-1)^8-(2x+3)^3\cdot8(4x^2-1)^7(4x^2-1)'}{(4x^2-1)^{16}}=
\frac{6(2x+3)^2(4x^2-1)^8-64x(2x+3)^3(4x^2-1)^7}{(4x^2-1)^{16}}=
\frac{6(2x+3)^2(4x^2-1)-64x(2x+3)^3}{(4x^2-1)^9}=
\frac{2(2x+3)^2(12x^2-3-64x^2-96x)}{(4x^2-1)^9}
=-\frac{2(2x+3)^2(52x^2+96x+3)}{(4x^2-1)^9}$$
