X & Y

The given equation can be reduced to the form :

$(xy) ^y=y^x$ . Hence $x$ must be an integral power of $y$ . Say $x=y^n$ where $n$ is a positive integer (since $x$ is a positive integer).

Then $y^{(n+1)y}=y^{y^n}\implies y^{n-1}=n+1$ . For $n\geq 4,y$ can never be an integer. So, let us consider the following cases :

$\boxed 1$ $n=0$

In this case, $x=y^0=1$ . Then $1^y=y^{1-y}\implies y^{1-y}=1\implies y=1$ .

$\boxed 2$ $n=1$

In this case, $y^y=y^0=1\implies y=1=x$ .

$\boxed 3$ $n=2$

In this case, $y=3\implies x=3^2=9$ .

$\boxed 4$ $n=3$

In this case, $y^2=4\implies y=2,x=2^3=8$ .

Hence, the sum of all the values of $x$ and $y$ is $1+1+9+3+8+2=\boxed {24}$ .