Try Setting $\frac xy$ As The Subject?

$\begin{aligned} \frac {x+y}{x-y} & = \sqrt 2 \\ x+y & = \sqrt 2 (x-y) \\ (x+y)^2 & = 2 (x-y)^2 \\ x^2+2xy+y^2 & = 2x^2-4xy+2y^2 \\ 2xy & = x^2-4xy+y^2 \\ x^2 + y^2 & = 6xy \\ \implies \frac {x^2+y^2}{xy} & = \frac {6xy}{xy} \\ & = \boxed{6} \quad \text{for } xy \ne 0 \end{aligned}$

First of all, note that;
If $x=0$ or $y=0$ ;
Then, $\frac{x+y}{x-y}$ will be $-1$ and $1$ respectively; So, we can safely conclude that $xy \neq 0$ ; hence the required fraction is NOT indeterminate for any value of the variables.

Now,
Let $\frac{x^2 + y^2}{xy} = a$ ;
$\implies \frac {x^2+y^2+2xy}{x^2+y^2-2xy} = \frac {a+2}{a-2}$ ;
$\implies \frac {(x+y)^2}{(x-y)^2} = \frac {a+2}{a-2}$ ;
$\implies 2 = \frac{a+2}{a-2}$ ;
$\implies 2a-4=a+2$ ;
$\implies \boxed{a=6}$

Note that componendo et dividendo was used in 2nd step and value of $\frac {x+2}{x-2}$ was put as $\sqrt {2}$ in 4th step and other steps follow.

$A = \frac {x+y}{x-y} = \sqrt {2}$

$x + y = \sqrt {2} x - \sqrt {2}y$

$x + y = \sqrt {2} x - \sqrt {2}y$

$( \sqrt {2} + 1) y = ( \sqrt {2} -1)x$

It is easy to see, that in the case of x=y=0, A would be undefined (division by zero); and in the cases ( x = 0 and y ≠ 0) or ( x ≠ 0 and y = 0), A = 1 would be true (instead of the A = square root of 2)).

Therefore, x≠ 0 and y ≠ 0 and we can divide by both x and y.

Now:

$\frac{x}{y} = \frac { \sqrt{2} + 1}{ \sqrt{2} - 1} = \frac { \sqrt{2} + 1}{ \sqrt{2} - 1} × \frac { \sqrt{2} + 1}{ \sqrt{2} + 1} = \frac {2 + 2 \sqrt{2} + 1}{ 2 - 1} = 3 + 2 \sqrt{2}$

Similarly:

$\frac{y}{x} = \frac { \sqrt{2} - 1}{ \sqrt{2} + 1} = \frac { \sqrt{2} - 1}{ \sqrt{2} + 1} × \frac { \sqrt{2} - 1}{ \sqrt{2} - 1} = \frac {2 - 2 \sqrt{2} + 1}{ 2 - 1} = 3 - 2 \sqrt{2}$

Then:

$B = \frac { x^2 + y^2}{xy} =\frac { x^2 }{xy} + \frac { y^2}{xy} = \frac {x}{y} + \frac {y}{x} = 3 + 2 \sqrt{2} + 3 - 2 \sqrt{2} = \boxed {6}$

we actually need to find $\frac{x}{y} +\frac {y}{x }$ from given data by dividing in numerator and denominator by $y$ and substiting value of $\frac{x}{y}$ .

First of all note that the condition for either of $x$ or $y$ to be zero is non-existent since, $\dfrac{x+y}{x-y} \neq \sqrt2 \$ for any value of the other non-zero variable if $xy=0$ .

So, proceeding by the hint as stated in the title of the problem, divide both the numerator and denominator of the LHS of the equation by $y$ to get,

$\dfrac{\frac xy +1}{\frac xy - 1} = \sqrt2$

Similarly for the expression to be obtained, one can write it as,

$\dfrac{{\left(\frac xy\right)}^2 +1}{\frac xy} \qquad \qquad \small\text{Dividing numerator and denominator by} \ y^2$

Take $\dfrac xy$ as a separate variable, $z$ ,and solve accordingly,

$\begin{aligned} & \dfrac{z+1}{z-1} = \sqrt2 \\ \implies & z = \dfrac{\sqrt2 +1}{\sqrt2 -1} = 3+2\sqrt2 \end{aligned}$

Plugging in $z$ in our required expression, we get,

$\dfrac{{\left(3+2\sqrt2\right)}^2 + 1}{3+2\sqrt2} = \boxed{6}$