#### Answer

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

#### Work Step by Step

$$xy=\cot(xy)$$
1) Differentiate both sides of the equation with respect to $x$:
$$\frac{d}{dx}(xy)=\frac{d}{dx}(\cot xy)$$ $$y+x\frac{dy}{dx}=-\csc^2(xy)\frac{d}{dx}(xy)$$ $$y+x\frac{dy}{dx}=-\csc^2(xy)\Big(y+x\frac{dy}{dx}\Big)$$ $$\Big(y+x\frac{dy}{dx}\Big)=-\csc^2(xy)y-\csc^2(xy)x\frac{dy}{dx}$$
2) Collect all the terms with $dy/dx$ onto one side and solve for $dy/dx$: $$x\frac{dy}{dx}+\csc^2(xy)x\frac{dy}{dx}=-\csc^2(xy)y-y$$ $$\frac{dy}{dx}(x+x\csc^2(xy))=-(y+y\csc^2(xy))$$ $$\frac{dy}{dx}=-\frac{y+y\csc^2(xy)}{x+x\csc^2(xy)}=-\frac{y(1+\csc^2(xy))}{x(1+\csc^2(xy))}=-\frac{y}{x}$$