Given the equation $\sqrt{x^{2}+2 x y+y^{6}}=8$, evaluate $\frac{d y}{d x}$. Assume that the equation implicitly defines $y$ as a differentiable function of $x$.
If $F(x, y)=\sqrt{x^{2}+2 x y+y^{6}}-8=0$, then $F_{x}=$
If $F(x, y)=\sqrt{x^{2}+2 x y+y^{6}}-8=0$, then $F_{y}=$
\[
\frac{d y}{d x}=
\]

Step 1 ：Given the equation \(\sqrt{x^{2}+2 x y+y^{6}}=8\), we are asked to evaluate \(\frac{d y}{d x}\). We assume that the equation implicitly defines \(y\) as a differentiable function of \(x\).

Step 2 ：We define a function \(F(x, y)=\sqrt{x^{2}+2 x y+y^{6}}-8=0\).

Step 3 ：We find the partial derivatives of \(F\) with respect to \(x\) and \(y\), denoted as \(F_{x}\) and \(F_{y}\) respectively.

Step 4 ：\(F_{x} = \frac{x + y}{\sqrt{x^{2} + 2*x*y + y^{6}}}\)

Step 5 ：\(F_{y} = \frac{x + 3*y^{5}}{\sqrt{x^{2} + 2*x*y + y^{6}}}\)

Step 6 ：We can find \(\frac{d y}{d x}\) using the implicit differentiation method, which gives us \(\frac{d y}{d x} = -\frac{F_{x}}{F_{y}}\).

Step 7 ：Substituting the values of \(F_{x}\) and \(F_{y}\) into the equation, we get \(\frac{d y}{d x} = -\frac{x + y}{x + 3y^{5}}\).

Step 8 ：\(\boxed{\frac{d y}{d x} = -\frac{x + y}{x + 3y^{5}}}\) is the final answer.