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

$d y-x=C(1+x y)$

We have,

$\frac{d y}{d x}=\frac{1+y^2}{1+x^2} \Rightarrow \frac{1}{1+y^2} d y=\frac{1}{1+x^2} d x$

On integrating, we get

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

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

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

$\Rightarrow \frac{y-x}{1+x y}=C \Rightarrow y-x=C(1+x y)$