Interesting Olympiad Math problem | Can you solve ?

[@ManojkantSamal]

Ans ::*3i/2...... May be
^=read as to the power
*=read as square root
Let (1/x)=y
So, xy=1
As per question
X+y=1.....eqn1
So,
X-y=*{(x+y)^2-4xy}
=*{1-4}=*(-3)=*3i..... Eqn2
Eqn1 +eqn2
X+y+x-y=1+*3i
So,
2x=1+*3i
X=(1+*3i)/2
So,
Y=(1/x)=2/(1+*3i)
={2(1-*3i)}/{(1+*3i)(1-*3i)}
={2(1-*3i)}/4
=(1-*3i)/2
Now, x^^2-y^2=
{(1+*3i)^2-(1+*3i)^2}/4
=(4×1×*3i)/4
=*3i.....May be
Another method
X^2-y^2=eqn1×eqn2
=1×(*3i)=*3i
So,
X^2-(1/x^2)=*3i

[@ChidexMath2-w4h]

Nice