The solution of the equation $(2+x) d y-y d x=0$ represents a curve passing through a fixed point $P$, then the area of equilateral triangle with $P$ as one vertex and $x+y=0$ as its one side, is

$\frac{2}{\sqrt{3}}$

We have,

$(2+x) d y-y d x=0 \Rightarrow \frac{1}{y} d y-\frac{1}{2+x} d x=0$

On integrating, we obtain

$\log y-\log (x+2)=\log C \Rightarrow y=C(x+2)$

Clearly, it represents a curve passing through a fixed point $P(-2,0)$.

The altitude of the equilateral triangle having one vertex at $P(-2,0)$ and opposite side $x+y=0$ is

$p=\left|\frac{-2+0}{\sqrt{1+1}}\right|=\sqrt{2}$

So, its area is $\frac{p^2}{\sqrt{3}}=\frac{2}{\sqrt{3}}$ sq. units