Step 1:

$y^2(2+x)=x^2(6-x)$

or $y^2=\large\frac{x^2(6-x)}{(2+x)}$

Step 2:

Existence : The curve is not defined for $x > 6$ and $x \leq -2$. so it lies in the interval $ -2 < x \leq 1$

Step 3:

(ii) Symmetry: The curve is symmetric about the x-axis only

Step 4:

(iii) Asymptotes: $x=-2$ is a vertical asymplote to the curve , parallel to the y-axis

Step 5:

The curve passes through $(0,0)$ twice $(6,0)$ once, and hence a loop is forward between $(0,0)\;and \; (6,0)$