Explanation

(a)
Since, to find $\ \ \frac{dy}{dx}\ at\ x = 0$
$\therefore At\ x = 0,{log\ }{(y + 0) = 0\ \ \ \ \ \ \ \ \Rightarrow \ \ \ y = 1}$
$\therefore To\ find\frac{dy}{dx}\ at\ (0,1)$
On differentiating given equation w.r.t. $x,\$we get
$\ \frac{1}{x + y}\left( 1 + \frac{dy}{dx} \right) = 2x\frac{dy}{dx} + 2y.1$
$\Rightarrow \ \ \frac{dy}{dx} = \frac{2y(x + y) - 1}{1 - 2(x + y)x}$
$\therefore\left( \frac{dy}{dx} \right)_{(0,1)} = 1$