${\tan ^{ - 1}}\left( {{x^2} + {y^2}} \right) = a$

We have, ${\tan ^{ - 1}}\left( {{x^2} + {y^2}} \right) = a$

On differentiating both sides w.r.t. $x$, we get
$\frac{d}{{dx}}{\tan ^{ - 1}}\left( {{x^2} + {y^2}} \right) = \frac{da}{{dx}}$

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

$\Rightarrow$ $2x + \frac{d}{{dy}}{y^2} \cdot \frac{{dy}}{{dx}} = 0$
$\Rightarrow$ $2y \cdot \frac{{dy}}{{dx}} = - 2x$

therefore,$\frac{{dy}}{{dx}} = \frac{{ - 2x}}{{2y}} = \frac{{ - x}}{y}$