There is an easier way

Observe that the formula is very similar to the distance formula, where the distance between $(x_1, y_1)$ and $(x_2, y_2)$ is equal to $\sqrt{ (x_1 - x_2) ^ 2 + ( y_1 - y_2) ^ 2 }$ .

If we consider the functions:
$\displaystyle{f\left( x \right) =\sqrt { 2-{ x }^{ 2 } } \Rightarrow f:\quad { x }^{ 2 }+{ y }^{ 2 }=2\\ g\left( x \right) =\cfrac { 9 }{ x } \Rightarrow g:\quad y=\cfrac { 9 }{ x } },$ and consider the points $(x_1, f(x_1) )$ and $(x_2, g(x_2) )$ , we see that our task is to minimize the distance between these two curves (and then square the value).

We know that the minimum distance between 2 continuous, differentiable curves must satisfy the condition that the line joining both their points of contact must be normal to both curves. With some work, we can show that this must be $y = x$ .

Solving, we get the points $(x_1, y_1) = (1,1)$ and $(x_2, y_2 ) = (3, 3 )$ .
Hence, the answer is equal to $(3-1)^2 + (3-1)^2 = 8$ .

By partial differentiation with respect to X1 we get. 2 (X1 - X2) = [ (rt(2 - X1^2)) - 9 / X2 ] 1 / (rt (2 - X1^2)) * X1 by partial differentiation with respect to X2 (X2 - X1) = [ 9 / X2 - (rt(2 - X1^2/00] * 9 / X2

equating both the equations, we get X1 * X2^2 / rt ( rt ( 2 - X1^2)) = +- 3

as X2 should be +ve real no, so X2 = 3 and X1 = 1

substitute X2 = 3 and X1 = 1 in given expression

=(-2)^2 + (1 - 3)^2

= 8