1x+ 3= y
eqn is
(y-3)^2 + \frac{9 ( y-3)^2}{y^2} = 16
(y-3)^2(y^2+9)= 16y^2
(y^2-6y+9)(y^2+9)= 16y^2
original :-
\frac{(y^2-6y+9)(y^2+9)}{y ( y) }= 16y^2
edit:- ON ADVICE OF NISHANT BHAIYYA
\frac{(y^2-6y+9)(y^2+9)}{y ( y) }= 16
\left ( y + \frac{9}{y} -6 \right )(y +\frac{9}{y}) =16
y + \frac{9}{y} = t\rightarrow (t-6)(t)= 16 \rightarrow t^2-6t-16 =0
we have
\boxed {t=8 } \, and \, \, \boxed{t=2}
we have
\boxed {y=4 \pm \sqrt{7}} \, and \, \, \boxed{y= -1 +i(2\sqrt{2})}
\boxed {x=1\pm \sqrt{7}} \, and \, \, \boxed{x= -4 \pm i(2\sqrt{2})}
