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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Differentiate the equation
$$
x^3 + x^2y - xy^3 = 4
$$
implicitly with respect to $$ x $$.

1. Differentiate each term:
   - For $$ x^3 $$:
     $$
     \frac{d}{dx}\left(x^3\right) = 3x^2.
     $$
   - For $$ x^2y $$ (using the product rule):
     $$
     \frac{d}{dx}\left(x^2y\right) = 2xy + x^2 \frac{dy}{dx}.
     $$
   - For $$-xy^3$$ (using the product rule):
     $$
     \frac{d}{dx}\left(-xy^3\right) = -\left(\frac{d}{dx}\left(x\right)y^3 + x\frac{d}{dx}\left(y^3\right)\right) = -\left(1\cdot y^3 + x\cdot 3y^2\frac{dy}{dx}\right) = -y^3 - 3xy^2\frac{dy}{dx}.
     $$

2. Write the differentiated equation:
   $$
   3x^2 + 2xy + x^2 \frac{dy}{dx} - y^3 - 3xy^2 \frac{dy}{dx} = 0.
   $$

3. Group the terms involving $$\frac{dy}{dx}$$:
   $$
   \left(x^2 - 3xy^2\right)\frac{dy}{dx} + 3x^2 + 2xy - y^3 = 0.
   $$

4. Isolate the $$\frac{dy}{dx}$$ term:
   $$
   \left(x^2 - 3xy^2\right)\frac{dy}{dx} = y^3 - 3x^2 - 2xy.
   $$

5. Finally, solve for $$\frac{dy}{dx}$$:
   $$
   \frac{dy}{dx} = \frac{y^3 - 3x^2 - 2xy}{x^2 - 3xy^2}.
   $$
$$ \frac{dy}{dx} = \frac{y^3 - 3x^2 - 2xy}{x^2 - 3xy^2} $$

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

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