$$ xy^2 = 1$$
From the equation, it can be said that it is a rectangular hyperbola . Now if I want to calculate the vertex of it , I do follows :
$$\rightarrow (\frac{x}{2}+\frac{y^2}{2})^2 - (\frac{x}{2} - \frac {y^2}{2})^2=1$$ $$\rightarrow X^2-Y^2=1$$
where, $X=\frac{x}{2}+\frac{y^2}{2}$ and $Y=\frac{x}{2}-\frac{y^2}{2}$ .
Now, for vertex : $$X=\pm a,Y=0 (a=1)$$ which implies : $$ \frac{x}{2}+\frac{y^2}{2}=\pm 1 ....(i) $$ $$ \frac{x}{2}-\frac{y^2}{2}=0 ....(ii) $$
Working with $(i)$ and $(ii)$ , I get : $x=\pm 2, y=0$ .
This, vertex is $(\pm 2,0)$ . Am I correct ?