Answer
$$ y'(x) = \frac{\sec ^{2} x-\sec ^{2}(x+y)}{\sec ^{2}(x+y)-\sec ^{2} y} $$
Work Step by Step
Given $$\tan (x+y)=\tan x+\tan y$$
Differentiate with respect to $x$
\begin{align*}
\frac{d}{dx} \tan (x+y)&=\frac{d}{dx}[\tan x+\tan y] \\
\sec^2 \left(x+ y\right)[1+y'(x) ]&=\sec^2 x+ \sec^2 y y'(x) \\
\left [\sec^2 \left(x+ y\right) -\sec^2 y \right] y'(x)& =\sec^2 x- \sec^2 \left(x+ y\right)
\end{align*}
Then
$$ y'(x) = \frac{\sec ^{2} x-\sec ^{2}(x+y)}{\sec ^{2}(x+y)-\sec ^{2} y} $$
