If $u = \log_e (x^2+y^2) + \tan^{-1} (y/x)$, then $\frac{\partial^2u}{dx^2} + \frac{\partial^2u}{\partial y^2}$ is equal to:

Question

( A ) $\frac{1}{2}u$
( B ) $2a$
( C ) $u$
( D ) $0$