Given: \[ y = (\sin x)^y. \] Take the natural logarithm of both sides: \[ \ln y = y \ln (\sin x). \] Differentiate both sides with respect to \( x \): \[ \frac{1}{y} \frac{dy}{dx} = \frac{d}{dx} \big[ y \ln (\sin x) \big]. \] Apply the product rule to the right-hand side: \[ \frac{1}{y} \frac{dy}{dx} = \frac{dy}{dx} \ln (\sin x) + y \frac{d}{dx} \big[ \ln (\sin x) \big]. \] The derivative of \( \ln (\sin x) \) is \( \frac{d}{dx} \ln (\sin x) = \cot x. \). Substitute this into the equation: \[ \frac{1}{y} \frac{dy}{dx} = \frac{dy}{dx} \ln (\sin x) + y \cot x. \] Multiply by \( y \) to remove the denominator: \[ \frac{dy}{dx} = y \frac{dy}{dx} \ln (\sin x) + y^2 \cot x. \] Rearrange to isolate \( \frac{dy}{dx} \): \[ \frac{dy}{dx} \big( 1 - y \ln (\sin x) \big) = y^2 \cot x. \] Solve for \( \frac{dy}{dx} \): \[ \frac{dy}{dx} = \frac{y^2 \cot x}{1 - y \ln (\sin x)}. \]
Final Answer: \[ \boxed{\frac{y^2 \cot x}{1 - y \ln (\sin x)}} \]