If $y=x\sin y$, then $\frac{dy}{dx}$ is:

$\frac{\sin^2 y}{\sin y-y \cos y}$

The correct answer is Option (2) → $\frac{\sin^2 y}{\sin y-y \cos y}$

$y = x \sin y$

Differentiate both sides:

$\frac{dy}{dx} = \sin y + x \cos y \cdot \frac{dy}{dx}$

Bring $\frac{dy}{dx}$ terms together:

$\frac{dy}{dx} - x \cos y \frac{dy}{dx} = \sin y$

$\frac{dy}{dx}(1 - x \cos y) = \sin y$

Since $x = \frac{y}{\sin y}$ from the original equation:

$1 - x \cos y = 1 - \frac{y \cos y}{\sin y} = \frac{\sin y - y \cos y}{\sin y}$

Hence:

$\frac{dy}{dx} = \frac{\sin y}{\frac{\sin y - y \cos y}{\sin y}} = \frac{\sin^2 y}{\sin y - y \cos y}$

$\frac{dy}{dx} = \frac{\sin^2 y}{\,\sin y - y \cos y\,}$