$$ 1 \leftarrow \quad $$ Given 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 function of $$ x $$. If $$ F(x, y)=\sqrt{x^{2}+4 x y+y^{6}}-8=0 $$, then $$ F_{x}=\square $$

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To find $$ \frac{dy}{dx} $$ for the equation $$ \sqrt{x^{2}+4xy+y^{6}}=8 $$, we will use implicit differentiation.

First, we define the function $$ F(x, y) $$ as follows:

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

We need to find the partial derivative $$ F_x $$ with respect to $$ x $$.

### Step 1: Calculate $$ F_x $$

Using the chain rule, we differentiate $$ F(x, y) $$:

$$
F_x = \frac{\partial}{\partial x} \left( \sqrt{x^{2} + 4xy + y^{6}} \right)
$$

Using the chain rule, we have:

$$
F_x = \frac{1}{2\sqrt{x^{2} + 4xy + y^{6}}} \cdot (2x + 4y + 0) = \frac{x + 2y}{\sqrt{x^{2} + 4xy + y^{6}}}
$$

### Step 2: Calculate $$ F_y $$

Next, we find the partial derivative $$ F_y $$:

$$
F_y = \frac{\partial}{\partial y} \left( \sqrt{x^{2} + 4xy + y^{6}} \right)
$$

Again using the chain rule:

$$
F_y = \frac{1}{2\sqrt{x^{2} + 4xy + y^{6}}} \cdot (4x + 6y^{5}) = \frac{2x + 3y^{5}}{\sqrt{x^{2} + 4xy + y^{6}}}
$$

### Step 3: Implicit Differentiation

Now we apply implicit differentiation to the equation $$ F(x, y) = 0 $$:

$$
\frac{dF}{dx} = F_x + F_y \frac{dy}{dx} = 0
$$

Rearranging gives:

$$
F_y \frac{dy}{dx} = -F_x
$$

Thus, we can solve for $$ \frac{dy}{dx} $$:

$$
\frac{dy}{dx} = -\frac{F_x}{F_y}
$$

### Step 4: Substitute $$ F_x $$ and $$ F_y $$

Now we substitute $$ F_x $$ and $$ F_y $$:

$$
\frac{dy}{dx} = -\frac{\frac{x + 2y}{\sqrt{x^{2} + 4xy + y^{6}}}}{\frac{2x + 3y^{5}}{\sqrt{x^{2} + 4xy + y^{6}}}} = -\frac{x + 2y}{2x + 3y^{5}}
$$

### Conclusion

Thus, the derivative $$ \frac{dy}{dx} $$ is:

$$
\frac{dy}{dx} = -\frac{x + 2y}{2x + 3y^{5}}
$$

Now, we can fill in the blank for $$ F_x $$:

$$
F_x = \frac{x + 2y}{\sqrt{x^{2} + 4xy + y^{6}}}
$$
$$ F_x = \frac{x + 2y}{\sqrt{x^{2} + 4xy + y^{6}}} $$

\( 1 \leftarrow \quad \) Given 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 function of \( x \). If \( F(x, y)=\sqrt{x^{2}+4 x y+y^{6}}-8=0 \), then \( F_{x}=\square \)

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