SOLUTION — At $x=0$,
$\log (0+y)=2 \times 0 \times y$
$\Rightarrow y=1$
On differentiating given equation, we get
$\begin{aligned}\frac{1}{x+y}\left(1+\frac{d y}{d x}\right) =2 x \frac{d y}{d x}+2 y \cdot 1 \\\Rightarrow \quad \frac{d y}{d x}=\frac{2 y(x+y)-1}{1-2(x+y) \cdot x} \Rightarrow\left(\frac{d y}{d x}\right)_{(0,1)} & =1\end{aligned}$
So, The correct option will be (A).