Answer
$$\frac{dy}{dx}=-\frac{y\sec^2(xy)+1}{x\sec^2(xy)}$$
Work Step by Step
$$x+\tan(xy)=0$$
1) Differentiate both sides of the equation with respect to $x$:
$$\frac{dx}{dx}+\frac{d}{dx}(\tan(xy))=0$$ $$1+\sec^2(xy)\frac{d}{dx}(xy)=0$$ $$1+\sec^2(xy)(y+x\frac{dy}{dx})=0$$ $$1+y\sec^2(xy)+x\sec^2(xy)\frac{dy}{dx}=0$$
2) Collect all the terms with $dy/dx$ onto one side and solve for $dy/dx$: $$x\sec^2(xy)\frac{dy}{dx}=-(y\sec^2(xy)+1)$$ $$\frac{dy}{dx}=-\frac{y\sec^2(xy)+1}{x\sec^2(xy)}$$
