Given equation can be rewritten as
$y=2+\sqrt{\sin x+y}$
$\Rightarrow {\left(y-2\right)}^2=\sin x+y$
$\Rightarrow y^2-4y+4=\sin x+y\dots \left(\mathrm{i}\right)$
Putting $x=0$ in the equation $\left(\mathrm{i}\right)$, we get,
$y^2-4y+4=0+y\Rightarrow y^2-5y+4=0$
$\Rightarrow \left(y-1\right)\left(y-4\right)=0$
$y=1$ or $y=4$
$\because y>2\Rightarrow y=4$
Now differentiating equation $\left(\mathrm{i}\right)$ with respect to $x$, we get,
$\frac{dy}{dx}=\frac{\cos x}{2y-5}$
Putting $x=0$, $y=4$

$\frac{dy}{dx}=\frac{\cos \left(0\right)}{2\left(4\right)-5}=\frac{1}{3}$