$${\left(\frac{2^x+2^{-x}}{2}\right)}^2-{\left(\frac{2^x-2^{-x}}{2}\right)}^2$$

$$={\left(\frac{2^x+{\displaystyle \frac{1}{2^x}}}{2}\right)}^2-{\left(\frac{2^x-{\displaystyle \frac{1}{2^x}}}{2}\right)}^2$$

$$={\left(\frac{{\displaystyle \frac{2^x}{1}}·{\displaystyle \frac{2^x}{2^x}}+{\displaystyle \frac{1}{2^x}}}{2}\right)}^2-{\left(\frac{{\displaystyle \frac{2^x}{1}}·{\displaystyle \frac{2^x}{2^x}}-{\displaystyle \frac{1}{2^x}}}{2}\right)}^2$$

$$={\left(\frac{{\displaystyle \frac{2^{2x}+1}{2^x}}}{{\displaystyle \frac{2}{1}}}\right)}^2-{\left(\frac{{\displaystyle \frac{2^x-1}{2^x}}}{{\displaystyle \frac{2}{1}}}\right)}^2$$

$$={\left(\frac{\left(2^{2x}+1\right)·1}{2^x·2}\right)}^2-{\left(\frac{\left(2^{2x}-1\right)·1}{2^x·2}\right)}^2$$

$$={\left(\frac{2^{2x}+1}{2^{x+1}}\right)}^2-{\left(\frac{2^{2x}-1}{2^{x+1}}\right)}^2$$

$$={\left(\frac{{\left(2·2\right)}^x+1}{2^{x+1}}\right)}^2-{\left(\frac{{\left(2·2\right)}^x-1}{2^{x+1}}\right)}^2$$

$$={\left(\frac{4^x+1}{2^{x+1}}\right)}^2-{\left(\frac{4^x-1}{2^{x+1}}\right)}^2$$

$$=\frac{{\left(4^x+1\right)}^2}{{\left(2^{x+1}\right)}^2}-\frac{{\left(4^x-1\right)}^2}{{\left(2^{x+1}\right)}^2}$$

$$=\frac{{16}^x+2·4^x·1+1}{2^{2x+2}}-\frac{{16}^x-2·4^x·1+1}{2^{2x+2}}$$

$$=\frac{{16}^x+2·2^{2x}·1+1}{2^{2x+2}}-\frac{{16}^x-2·2^{2x}·1+1}{2^{2x+2}}$$

$$=\frac{{16}^x+2^{2x+1}·1+1}{2^{2x+2}}-\frac{{16}^x-2^{2x+1}·1+1}{2^{2x+2}}$$

$$=\frac{\cancel{{16}^x}+2^{2x+1}+\bcancel{1}-\cancel{{16}^x}+2^{2x+1}-\bcancel{1}}{2^{2x+2}}$$

$$=\frac{2^{2x+1}+2^{2x+1}}{2^{2x+2}}$$

$$=\frac{2·2^{2x+1}}{2^{2x+2}}$$

$$=\frac{2^{2x+2}}{2^{2x+2}}$$