Given, $\log \left(x+y\right)-2xy=0 .........\left(1\right)$

Differentiate equation $\left(1\right)$ w.r.t $x$

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

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

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

Hence,$y^{\text{'}}\left(0\right)=-\frac{1-0-2{y_{\left(0\right)}}^2}{1-0-0} ........\left(2\right)$

Now put $x=0$ in equation $\left(1\right)$

$\log \left(0+y\right)=0\Rightarrow y_{\left(0\right)}=1$

from equation $\left(2\right)$ 

$y^{\text{'}}\left(0\right)=1$