10. [-/1.07 Points] DETAILS MY NC Find $$ \frac{d y}{d x} $$ by implicit differentiation. $$ \frac{d y}{d x}=\square x^{3}+x^{2} y-x y^{3}=4 $$

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To find $$ \frac{d y}{d x} $$ by implicit differentiation for the equation $$ x^{3} + x^{2} y - x y^{3} = 4 $$, we will follow these steps:

1. Differentiate both sides of the equation with respect to $$ x $$.
2. Apply the product rule where necessary.
3. Solve for $$ \frac{d y}{d x} $$.

Let's differentiate the equation step by step.

1. Differentiate $$ x^{3} $$:
   $$
   \frac{d}{dx}(x^{3}) = 3x^{2}
   $$

2. Differentiate $$ x^{2} y $$ using the product rule:
   $$
   \frac{d}{dx}(x^{2} y) = x^{2} \frac{d y}{d x} + 2xy
   $$

3. Differentiate $$ -x y^{3} $$ using the product rule:
   $$
   \frac{d}{dx}(-x y^{3}) = -y^{3} - 3xy^{2} \frac{d y}{d x}
   $$

4. Differentiate the constant $$ 4 $$:
   $$
   \frac{d}{dx}(4) = 0
   $$

Now, we can combine these results into the differentiated equation:
$$
3x^{2} + \left( x^{2} \frac{d y}{d x} + 2xy \right) - \left( y^{3} + 3xy^{2} \frac{d y}{d x} \right) = 0
$$

Now, let's rearrange this equation to solve for $$ \frac{d y}{d x} $$.

Combining like terms gives us:
$$
3x^{2} + 2xy - y^{3} + x^{2} \frac{d y}{d x} - 3xy^{2} \frac{d y}{d x} = 0
$$

Now, we can isolate $$ \frac{d y}{d x} $$:
$$
x^{2} \frac{d y}{d x} - 3xy^{2} \frac{d y}{d x} = -3x^{2} - 2xy + y^{3}
$$

Factoring out $$ \frac{d y}{d x} $$:
$$
\frac{d y}{d x} (x^{2} - 3xy^{2}) = -3x^{2} - 2xy + y^{3}
$$

Finally, we can solve for $$ \frac{d y}{d x} $$:
$$
\frac{d y}{d x} = \frac{-3x^{2} - 2xy + y^{3}}{x^{2} - 3xy^{2}}
$$

Thus, the result is:
$$
\frac{d y}{d x} = \frac{-3x^{2} - 2xy + y^{3}}{x^{2} - 3xy^{2}}
$$
$$ \frac{d y}{d x} = \frac{-3x^{2} - 2xy + y^{3}}{x^{2} - 3xy^{2}} $$

[-/1.07 Points] DETAILS MY NC
Find by implicit differentiation.

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[-/1.07 Points] DETAILS MY NC Find by implicit differentiation.