Answer
$\sqrt[4]{ 2^{}x^{}y^2}$
Work Step by Step
Using $a^{-x}=\dfrac{1}{a^x}$ or $\dfrac{1}{a^{-x}}=a^x,$ then
\begin{array}{l}\require{cancel}
\sqrt[4]{\sqrt[3]{8x^3y^6}}
\\\\=
\sqrt[4]{ \left( 8x^3y^6 \right)^{1/3}}
\\\\=
\left( \left( 8x^3y^6 \right)^{1/3} \right)^{1/4}
.\end{array}
Using the Product Rule of the laws of exponents which is given by $x^m\cdot x^n=x^{m+n},$ the expression above is equivalent to
\begin{array}{l}\require{cancel}
\left( \left( 8x^3y^6 \right)^{1/3} \right)^{1/4}
\\\\=
\left( 8x^3y^6 \right)^{\frac{1}{3}\cdot\frac{1}{4}}
\\\\=
\left( 8x^3y^6 \right)^{\frac{1}{12}}
\\\\=
\left( 2^3x^3y^6 \right)^{\frac{1}{12}}
.\end{array}
Using the extended Power Rule of the laws of exponents which is given by $\left( x^my^n \right)^p=x^{mp}y^{np},$ the expression above is equivalent to
\begin{array}{l}\require{cancel}
\left( 2^3x^3y^6 \right)^{\frac{1}{12}}
\\\\=
2^{3\cdot\frac{1}{12}}x^{3\cdot\frac{1}{12}}y^{6\cdot\frac{1}{12}}
\\\\=
2^{\frac{1}{4}}x^{\cdot\frac{1}{4}}y^{\frac{2}{4}}
.\end{array}
Using $x^{m/n}=\sqrt[n]{x^m}=\left(\sqrt[n]{x} \right)^m,$ then
\begin{array}{l}\require{cancel}
2^{\frac{1}{4}}x^{\cdot\frac{1}{4}}y^{\frac{2}{4}}
\\\\=
\sqrt[4]{ 2^{1}x^{1}y^2}
\\\\=
\sqrt[4]{ 2^{}x^{}y^2}
.\end{array}
