Answer
$\dfrac{dy}{dx}=\dfrac{1-\sec^2{(x+y)}}{\sec^2{(x+y)}}.$
At $(0, 0)\rightarrow\dfrac{dy}{dx}=\dfrac{1-\sec^2{(0+0)}}{\sec^2{(0+0)}}=0.$
Work Step by Step
$\dfrac{d}{dx}(\tan{(x+y)})=\dfrac{d}{dx}(x)\rightarrow$
Using the Chain Rule with $u=x+y\rightarrow\dfrac{du}{dx}=1+\dfrac{dy}{dx}\rightarrow$
$\dfrac{d}{dx}(\tan{(x+y)})=(1+\dfrac{dy}{dx})\sec^2{(x+y)}\rightarrow$
$(1+\dfrac{dy}{dx})\sec^2{(x+y)}=1\rightarrow$
$\dfrac{dy}{dx}=\dfrac{1-\sec^2{(x+y)}}{\sec^2{(x+y)}}.$
At $(0, 0)\rightarrow\dfrac{dy}{dx}=\dfrac{1-\sec^2{(0+0)}}{\sec^2{(0+0)}}=0.$
