Answer
$$\frac{dy}{dx}=\frac{-y+\sec^2(x+y)}{x-\sec^2(x+y))}.$$
Work Step by Step
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
$$\frac{dy}{dx}=\frac{-y+\sec^2(x+y)}{x-\sec^2(x+y))}.$$
