If y is a function of x and ${\log }_e\left(x+y\right)-2xy=0$ , then  the value of $y^1\left(0\right)$  is

${\log }_e\left(x+y\right)=2xy$

Differentiating with respect to x

$\frac{1}{x+y}\left\{1+\frac{dy}{dx}\right\}=2\left[x\frac{dy}{dx}+y\right]$

$x=0,y=1$

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

$\Rightarrow \frac{dy}{dx}=1$

so $y^1\left(x\right)$  is constant function

$\therefore y^1\left(0\right)=1$