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 $$ F_{y}=\square $$

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We start with the function

$$
F(x, y)=\sqrt{x^{2}+4xy+y^{6}}-8=0.
$$

The partial derivative with respect to $$y$$ is computed by applying the chain rule. Let

$$
u=x^{2}+4xy+y^{6}.
$$

Then

$$
F(x, y)=\sqrt{u}-8.
$$

Taking the derivative with respect to $$y$$:

$$
F_y=\frac{d}{dy}\left(\sqrt{u}\right)
=\frac{1}{2\sqrt{u}}\,\frac{du}{dy}.
$$

Now, we find $$\frac{du}{dy}$$. Differentiating $$u$$ with respect to $$y$$:

$$
\frac{du}{dy}=0+4x+6y^{5}=4x+6y^{5}.
$$

Therefore, we have

$$
F_y=\frac{4x+6y^{5}}{2\sqrt{x^{2}+4xy+y^{6}}}
=\frac{2x+3y^{5}}{\sqrt{x^{2}+4xy+y^{6}}}.
$$
$$ F_y = \frac{2x + 3y^{5}}{\sqrt{x^{2} + 4xy + y^{6}}} $$

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