1. **Problem:** Evaluate $\frac{dy}{dx}$ when $1 + xy^2 + x^2y = 0$. 2. **Formula and rules:** Use implicit differentiation. Differentiate both sides of the equation with respect to $x$. Remember to use the product rule: $\frac{d}{dx}[uv] = u'v + uv'$. 3. Differentiate each term: - $\frac{d}{dx}[1] = 0$ - $\frac{d}{dx}[xy^2] = y^2 + x \cdot 2y \frac{dy}{dx} = y^2 + 2xy \frac{dy}{dx}$ - $\frac{d}{dx}[x^2y] = 2xy + x^2 \frac{dy}{dx}$ 4. Substitute into the differentiated equation: $$0 + y^2 + 2xy \frac{dy}{dx} + 2xy + x^2 \frac{dy}{dx} = 0$$ 5. Group terms with $\frac{dy}{dx}$ and without: $$y^2 + 2xy + (2xy + x^2) \frac{dy}{dx} = 0$$ 6. Isolate $\frac{dy}{dx}$: $$ (2xy + x^2) \frac{dy}{dx} = - (y^2 + 2xy) $$ 7. Solve for $\frac{dy}{dx}$: $$ \frac{dy}{dx} = \frac{-(y^2 + 2xy)}{2xy + x^2} = \frac{-y(y + 2x)}{x(2y + x)} $$ 8. Note that $x(2y + x) = x(x + 2y)$ and the denominator can be written as $x(x + y)$ only if terms match. Here, the denominator is $x(2y + x)$ which is $x(x + 2y)$. 9. The closest matching option is (B): $\frac{dy}{dx} = -\frac{y(y + 2x)}{x(x + y)}$. **Final answer:** (B) $-\frac{y(y + 2x)}{x(x + y)}$