The equation in the image is:

$(2^{-x})^x = x^y$

Let's solve for $x$ and $y$.

Step 1: Simplify the left-hand side of the equation

$(2^{-x})^x = 2^{-x \cdot x} = 2^{-x^2}$ Thus, the equation becomes: $2^{-x^2} = x^y$

Step 2: Consider possible values of $x$ and $y$

This equation is not straightforward to solve analytically without further context or constraints on $x$ and $y$. However, we can consider specific cases for $x$ and $y$ to see if they satisfy the equation.

Case 1: $x = 1$

$2^{-1^2} = 1^y \implies 2^{-1} = 1 \implies \frac{1}{2} = 1$ This is not true, so $x = 1$ is not a solution.

Case 2: $x = 2$

$2^{-2^2} = 2^y \implies 2^{-4} = 2^y \implies 2^y = \frac{1}{16}$ Thus, $y = -4$.

Therefore, one solution is $x = 2$ and $y = -4$.

Conclusion

The solution to the equation