$\underset{x\to 0}{Lt}\left(\left(\frac{1-{\cos }^2\left(3x\right)}{{\cos }^3\left(4x\right)}\right)\left(\frac{{\sin }^{3}4x}{{\left(\log (2x+1)\right)}^5}\right)\right)$ is equal to

$\underset{x\to 0}{Lt}\left(\frac{1-{\cos }^{2}3x}{9x^2}\right)\left(\frac{9x^2}{{\cos }^{3}4x}\right).\frac{{\left(\frac{\sin 4x}{4x}\right)}^3\times 64x^3}{{\left(\frac{\log (1+2x)}{2x}\right)}^{5}32x^5}$

$\underset{x\to 0}{Lt}\frac{{\sin }^{2}3x}{{\left(3x\right)}^2}\frac{9}{{\cos }^{3}4x}.\frac{{\left(\frac{\sin 4x}{4x}\right)}^3.64}{{\left[\log {(1+2x)}^{\frac{1}{2x}}\right]}^5}\frac{1}{32}={\left(1\right)}^2.\frac{9}{{\left(1\right)}^3}.\frac{1^3\left(2\right)}{1^5}=18$