${\tan ^{ - 1}}x + {\tan ^{ - 1}}y = C$ is general Solution of the differential equation

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

On differentiating w.r.t. $x$,

we get
$\frac{1}{{1 + {x^2}}} + \frac{1}{{1 + {y^2}}} \cdot \frac{{dy}}{{dx}} = 0$

$\Rightarrow$ $\frac{1}{{1 + {y^2}}} \cdot \frac{{dy}}{{dx}} = - \frac{1}{{1 + {x^2}}}$

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