$\frac{1}{16}(\frac{4}{3}x-1)(\frac{4}{3}x+1)(\frac{16}{9}{x}^2+1)+0.5(\frac{4}{9}{x}^2+0.25)-{(\frac{4}{9}{x}^2+0.25)}^2=$

$=\frac{1}{16}(\frac{16}{9}{x}^2-1)(\frac{16}{9}{x}^2+1)+\frac{1}{2}(\frac{4}{9}{x}^2+\frac{1}{4})-(\frac{16}{81}{x}^4+\frac{2}{9}{x}^2+\frac{1}{16})=$

$=\frac{1}{16}(\frac{256}{81}{x}^4-1)+\frac{2}{9}{x}^2+\frac{1}{8}-\frac{16}{81}{x}^4-\frac{2}{9}{x}^2-\frac{1}{16}=$

$=\frac{16}{81}{x}^4-\frac{1}{16}+\frac{2}{9}{x}^2+\frac{1}{8}-\frac{16}{81}{x}^4-\frac{2}{9}{x}^2-\frac{1}{16}$= 0