jmc

In general, $$1-x+x^2=\frac{1+x^3}{1+x}.$$Thus, $$\begin{array}{rl}\prod _{n=0}^{\infty }\left[1-{\left(\frac{1}{2}\right)}^{3^n}+{\left(\frac{1}{4}\right)}^{3^n}\right]& =\prod _{n=0}^{\infty }\frac{1+{\left(\frac{1}{2}\right)}^{3^{n+1}}}{1+{\left(\frac{1}{2}\right)}^{3^n}}\\ & =\frac{1+{\left(\frac{1}{2}\right)}^3}{1+{\left(\frac{1}{2}\right)}^0}\cdot \frac{1+{\left(\frac{1}{2}\right)}^{3^2}}{1+{\left(\frac{1}{2}\right)}^3}\cdot \frac{1+{\left(\frac{1}{2}\right)}^{3^3}}{1+{\left(\frac{1}{2}\right)}^{3^2}}\cdots \\ & =\frac{1}{1+\frac{1}{2}}=\boxed{\frac{2}{3}}.\end{array}$$