\lim_{x\to 0}{\frac{1-\cos^\frac{4}{5} {3x}}{x^2}=\lim_{x\to 0} \frac{(1+(\cos {3x}-1))^\frac{4}{5}-1}{\cos {3x}-1}\cdot \frac{1-\cos {3x}}{(3x)^2}\cdot 9=\frac{4}{5}\cdot \frac{9}{2}=\frac{18}{5}