The solution of the differential equation $dy={\tan }^2\left(x+y\right)dx$

$\begin{array}{l}\frac{dy}{dx}={\tan }^2(x+y)\\ \text{ Put }x+y=t\Rightarrow 1+\frac{dy}{dx}=\frac{dt}{dx}\\ \frac{dt}{dx}-1={\tan }^{2}t\\ \Rightarrow \frac{dt}{dx}=1+{\tan }^{2}t={\sec }^{2}t\\ \Rightarrow {\cos }^{2}tdt=dx\\ \Rightarrow \int \frac{1+\cos 2t}{2}dt=\int dx\\ \Rightarrow \frac{1}{2}\left(t+\frac{\sin 2t}{2}\right)=x+c\end{array}$

$\begin{array}{l}\Rightarrow \frac{1}{4}\left[2\right(x+y)+\sin 2(x+y\left)\right]=x+c\\ \Rightarrow \sin (2x+2y)=2x-2y+c\end{array}$