Siven the equation \( \sqrt{x^{2}+4 x y+y^{6}}=8 \), evaluate \( \frac{d y}{d x} \). Assume that the equation implicitly defines \( y \) as a differentiable \( F(x, y)=\sqrt{x^{2}+4 x y+y^{6}}-8=0 \), then \( F_{x}=\frac{x+2 y}{\sqrt{x^{2}+4 x y+y^{6}}} \) \( F(x, y)=\sqrt{x^{2}+4 x y+y^{6}}-8=0 \), then \( F_{y}=\square \)

To find \( F_y \), we need to differentiate \( F(x, y) \) with respect to \( y \). The function \( F(x, y) \) is defined as: \[ F(x, y) = \sqrt{x^2 + 4xy + y^6} - 8 \] Now, taking the derivative of \( F \) with respect to \( y \): \[ F_y = \frac{\partial}{\partial y} \left( \sqrt{x^2 + 4xy + y^6} \right) \] Using the chain rule: \[ F_y = \frac{1}{2\sqrt{x^2 + 4xy + y^6}} \cdot (4x + 6y^5) \] Hence, we can summarize this as: \[ F_y = \frac{4x + 6y^5}{2\sqrt{x^2 + 4xy + y^6}} = \frac{2x + 3y^5}{\sqrt{x^2 + 4xy + y^6}} \] So, \( F_y \) evaluates to \( \frac{2x + 3y^5}{\sqrt{x^2 + 4xy + y^6}} \).