(a)
Differentiate: \( x^2 + y^2 = 1 \)
Left: \( \frac{d}{dx}(x^2 + y^2) = 2x + 2y \frac{dy}{dx} \)
Right: \( \frac{d}{dx}(1) = 0 \)
Equate: \( 2x + 2y \frac{dy}{dx} = 0 \)
Solve: \( \frac{dy}{dx} = -\frac{2x}{2y} = -\frac{x}{y} \)
Result: \( \frac{dy}{dx} = -\frac{x}{y} \) [4]

(b)
Given: \( \frac{dy}{dx} = -\frac{x}{y} \)
Separate variables: \( y \, dy = -x \, dx \)
Integrate: \( \int y \, dy = \int -x \, dx \implies \frac{y^2}{2} = -\frac{x^2}{2} + c \)
Multiply by 2: \( x^2 + y^2 = 2c = C \)
Initial condition: \( x = 1 \), \( y = 0 \implies 1^2 + 0^2 = C \implies C = 1 \)
Equation: \( x^2 + y^2 = 1 \)
Verify: Differentiate \( x^2 + y^2 = 1 \implies 2x + 2y \frac{dy}{dx} = 0 \implies \frac{dy}{dx} = -\frac{x}{y} \)
Result: \( x^2 + y^2 = 1 \) [5]