playing with x and y

$\left( { y }^{ 2 }+\frac { 1 }{ { y }^{ 2 } } -2 \right) +\left( { x }^{ 2 }+\frac { 1 }{ { x }^{ 2 } } -2 \right) =0$

${ \left( x-\frac { 1 }{ x } \right) }^{ 2 }+{ \left( y-\frac { 1 }{ y } \right) }^{ 2 }=0$

$x=\frac { 1 }{ x } \quad \therefore { \quad x }^{ 2 }=1$

$y=\frac { 1 }{ y } \quad \therefore { \quad y }^{ 2 }=1$

${ x }^{ 2 }+{ y }^{ 2 }=1+1=\boxed { 2 }$

I am assuming that the intended equation is $x^{2} + y^{2} + \frac{1}{x^{2}} + \frac{1}{y^{2}} = 4$ , for the equation as presently stated would not render a unique value for $x^{2} + y^{2}$ .

By the AM-GM inequality we have that

$x^{2} + \frac{1}{x^{2}} \ge 2$ ,

with equality only holding for $x = \pm 1$ .

The same holds true for $y^2 + \frac{1}{y^{2}}$ , and so

$x^{2} + y^{2} + \frac{1}{x^{2}} + \frac{1}{y^{2}} \ge 4$ ,

with equality only holding for $x = \pm 1, y = \pm 1$ .

For every possible option for $x, y$ we always have $x^{2} + y^{2} = 1 + 1 = \boxed{2}$ .

We have $x^2+y^2+\frac{1}{x^2}+\frac{1}{x^2}=4$

so $x^2+y^2+\frac{2}{x^2}=4$

then we multiply the equation by $x^2$ to get rid of the fraction

so we get $x^4+x^2y^2+2=4$

we get rid of the 2

now we have $x^4+x^2y^2=2$

Then we divide the equation by $x^2$ to get back to a second degree equation

$x^2+y^2=\frac{2}{x^2}$

now we substitute the $x^2+y^2$ with $\frac{2}{x^2}$

$\frac{2}{x^2} +\frac{1+1}{x^2}=4$

and now we have

$\frac{4}{x^2}=4$

now it's obvious $4x^2=4$

so $x^2 = 1$ and ofc $\boxed{x = 1}$

now simply substitute the x with its quantity in the first equation (1)

$1^2+y^2+\frac{1}{1^2}+\frac{1}{1^2}=4$

$3+y^2=4$

$y^2=1$ $\boxed{y=1}$

so $\boxed{x^2+y^2=1+1=\boxed{2}}$