[tex]\dfrac{\mathrm dy}{\mathrm dx}=\tan^2(x+y)[/tex]
Let [tex]v=x+y[/tex], so that [tex]\frac{\mathrm dv}{\mathrm dx}=1+\frac{\mathrm dy}{\mathrm dx}[/tex]:
[tex]\dfrac{\mathrm dv}{\mathrm dx}-1=\tan^2v[/tex]
Recall that [tex]\tan^2v+1=\sec^2v[/tex]:
[tex]\dfrac{\mathrm dv}{\mathrm dx}=\sec^2v[/tex]
Separate the variables and integrate both sides:
[tex]\cos^2v\,\mathrm dv=\mathrm dx[/tex]
[tex]\dfrac v2+\dfrac{\sin(2v)}4=x+C[/tex]
Solve in terms of [tex]y[/tex]:
[tex]\dfrac{x+y}2+\dfrac{\sin(2x+2y)}4=x+C[/tex]
