\(x^2=2^{64}\)
Can be written as
\(x^2=2^{32*2}\) = \({2^{32}}^2\) [ Using \(a^{b*c}\) = \((a^b)^c\) ]
Using "If base of two exponents is same then their power will also be same"
=> \(x = 2^{32}\) [ Using \(a^b\) = \(a^c\) => b = c

\(x^x=2^y\)
=> \(({2^{32}})^{2^{32}} =2^y\)
=> \(2^{{32*2^{32}}} =2^y\) [ Using \((a^b)^c\) = \(a^{b*c}\) ]
Now we know that 32 = \(2^5\)
=> \(2^{2^5*2^{32} =2^y\)
Using if two exponents with same base are multiplied then their power gets added up.
Now \(2^5\) * \(2^{32}\) will become \(2^{5 + 32}\) = \(2^{37}\) [Using \(x^a * x^b = x^{a+b}\) ]
=> \(2^{2^{37}} = 2^y\)
=> y = \(2^{37}\)

So, Answer will be D
Hope it helps!