## Calculus (3rd Edition)

$$\frac{dy}{dx}=\frac{-y+\sec^2(x+y)}{x-\sec^2(x+y))}.$$
Recall that $(\tan x)'=\sec^2 x$. By differentiating the equation $xy=\tan (x+y)$ with respect to $x$, we get $$y+x\frac{dy}{dx}=(1+\frac{dy}{dx})\sec^2(x+y)\\ \Longrightarrow \frac{dy}{dx}(x-\sec^2(x+y))=-y+\sec^2(x+y)$$ and hence $$\frac{dy}{dx}=\frac{-y+\sec^2(x+y)}{x-\sec^2(x+y))}.$$