Zaczynamy oczywiście od wykonania działań w nawiasach :).
[tex]\bigg[y^3:\Big(y^2\cdot y^{-3}\Big)\bigg]^{-4}:\left[\left(\frac1y\right)^4\cdot\frac1{y^{-2}}\right]^{-3}=\\\\\\=\bigg[\bigy^3:\big y^{2+(-3)}\bigg]^{-4}:\bigg[\big y^{-4}\cdot\big y^2\bigg]^{-3}=\\\\\\=\bigg[\big y^3:\big y^{-1}\bigg]^{-4}:\bigg[\big y^{-4+2}\bigg]^{-3}=\\\\\\=\bigg[\big y^{3-(-1)}\bigg]^{-4}:\bigg[\big y^{-2}\bigg]^{-3}=\\\\\\=\bigg[\big y^4\bigg]^{-4}:\big y^{-2\,\big\cdot(-3)}=\\\\\\=\big y^{4\,\big\cdot(-4)}:\big y^6=\\\\=\big y^{-16}:\big y^6=\\\\=\big y^{-16-6}=\\\\=\big y^{-22}[/tex]
Skoro [tex]y=\sqrt{2\sqrt2\,}[/tex] to:
[tex]\bigg[y^3:\Big(y^2\cdot y^{-3}\Big)\bigg]^{-4}:\left[\left(\frac1y\right)^4\cdot\frac1{y^{-2}}\right]^{-3}=\\\\\\=\big y^{-22}=\bigg(\sqrt{2\sqrt2\ }\bigg)^{-22}=\bigg(\sqrt{2\sqrt2\ }\bigg)^{2\cdot(-11)}=\bigg(2\sqrt2\,\bigg)^{-11}=\\\\\\=\bigg(\Big2^{11}\cdot\sqrt2\Big\,^{11}\bigg)^{-1}=\bigg(\Big2^{11}\cdot\sqrt2\Big\,^{2\cdot5+1}\bigg)^{-1}=\bigg(\Big2^{11}\cdot\Big(\sqrt2\big\, ^2\Big)^5\cdot\sqrt2\bigg)^{-1}=[/tex]
[tex]= \bigg(\Big2^{11}\cdot\Big2^5\cdot\sqrt2\bigg)^{-1}= \bigg(\Big2^{11+5}\cdot\sqrt2\bigg)^{-1}= \bigg(\Big2^{16}\cdot\sqrt2\bigg)^{-1}= \\\\\\=\bigg(65\,536\sqrt2\bigg)^{-1}=\dfrac1{65\,536\sqrt2}=\dfrac{\sqrt2}{13\,072}[/tex]