The solution of the differential equation $\left(1+x^2\right) \frac{d y}{d x}+1+y^2=0$, is

$\tan ^{-1} y+\tan ^{-1} x=\tan ^{-1} C$

If $\tan ^{-1} y+\tan ^{-1} x=\tan ^{-1} C$, then

$\frac{d}{d x}\left(\tan ^{-1} y\right)+\frac{d}{d x}\left(\tan ^{-1} x\right)=0$

$\Rightarrow \frac{1}{1+y^2} \frac{d y}{d x}+\frac{1}{1+x^2}=0 \Rightarrow\left(1+x^2\right) \frac{d y}{d x}+\frac{1}{1+y^2}=0$

Hence, $\tan ^{-1} x+\tan ^{-1} y=\tan ^{-1} C$ is the solution of the given differential.