A continuation

$y=\sum _{ n=1 }^{ x }{ \gcd{(x,n)} } = \gcd{(x,x)} + \sum _{ n=1 }^{ x -1}{ \gcd{(x,n)} } =x + \sum _{ n=1 }^{ x - 1 }{ \gcd{(x,n)} }$

Note that $y = 2x-1$ if $x$ is prime

Also note that if $x$ is composite, $\sum _{ n=1 }^{ x - 1 }{ \gcd{(x,n)} }$ is strictly greater than $x -1$ because there exists at least one $1 < n < x$ such that $gcd(x,n) > 1$

Hence, the only points lying on $y = 2x -1$ are points where $x$ is prime.

Hence, $D_{e^k} = (\text{Number of primes} \le e^k)$

By Prime Number Theorem, its limiting value when $k$ and hence $e^{k}$ tends to infinity is $\frac{e^{k}}{\ln{(e^{k})}} = \frac{e^{k}}{k}$

Hence, $(kD_{e^k})^{1/k} = (e^{k})^{1/k} = \boxed{e}$