Pretty simple, maybe?

Without loss of generality, let $y = tx$ , where $t$ is a real number. Then:

$\large \displaystyle x^{tx} = (tx)^x$

$\large \displaystyle x = (tx)^{\frac{1}{t}}$

$\large \displaystyle x = t^{\frac{1}{t}} \cdot x^{\frac{1}{t}}$

$\large \displaystyle x^{1 - \frac{1}{t}} = t^{\frac{1}{t}}$

$\color{#20A900} \boxed{ \large \displaystyle x = t^{\frac{1}{t-1}} }$

$\color{#20A900} \boxed{ \large \displaystyle y = tx = t^{\frac{t}{t-1}} }$

This is a family of solutions based in a parametric parameter $t$ . So we're looking to minimize:

$\large \displaystyle S = t^{\frac{1}{t-1}} + t^{\frac{t}{t-1}}$

Differtiating and making it equal to zero:

$\large \displaystyle \frac{dS}{dt} = t^{\frac{1}{t-1}} \left ( \frac{\frac{1}{t} (t-1) - \ln(t) }{(t-1)^2} \right ) + t^{\frac{t}{t-1}} \left ( \frac{(1 + \ln(t))(t-1) - t \ln(t)}{(t-1)^2} \right ) = 0$

Multiplying the first term above and below by $t$ :

$\large \displaystyle \frac{dS}{dt} = t^{\frac{t}{t-1}} \frac{1}{t} \left ( \frac{\frac{1}{t} (t-1) - \ln(t) }{(t-1)^2} \right ) + t^{\frac{t}{t-1}} \left ( \frac{(1 + \ln(t))(t-1) - t \ln(t)}{(t-1)^2} \right ) = 0$

Multiplying above and below by $\frac{ (t-1)^2 }{ t^{\frac{t}{t-1}} }$ (which implies $t \neq 0$ and $t \neq 1 \rightarrow x \neq y$ , as the problem states):

$\large \displaystyle \frac{1}{t} \left ( \frac{t-1}{t} - \ln(t) \right ) + (t-1)(1 + \ln(t)) - t\ln(t) = 0$

Multiplying above and below by $t^2$ :

$\large \displaystyle t^3 - t^2 - t^2 \ln(t) + t - t \ln(t) - 1 = 0$

The only solution to this is $t=1$ , which is forbidden by the steps above and by the problem itself ( $x \neq y$ ). This means we're looking for a limiting condition, that's never achieved. I.e., the minimum M never is attained, but is tends to the following value:

$\large \displaystyle M = \lim_{t \rightarrow 1} S$

$\large \displaystyle M = \lim_{t \rightarrow 1} \left [ t^{\frac{1}{t-1}} + t^{\frac{t}{t-1}} \right ]$

$\large \displaystyle M = \lim_{t \rightarrow 1} \left [ t^{\frac{t}{t-1}} \left ( \frac{1}{t} + 1 \right ) \right ]$

$\large \displaystyle \ln(M) = \lim_{t \rightarrow 1} \left [ \ln \left ( \frac{1}{t} + 1 \right ) + \frac{t}{t-1} \ln(t) \right ]$

$\large \displaystyle \ln(M) = \ln(2) + \lim_{t \rightarrow 1} \left [ \frac{t \ln(t)}{t-1} \right ]$

L'Hôpital's rule applies:

$\large \displaystyle \ln(M) = \ln(2) + \lim_{t \rightarrow 1} \left [ \frac{1 + \ln(t)}{1} \right ]$

$\large \displaystyle \ln(M) = \ln(2) + 1$

$\color{#3D99F6} \boxed{ \large \displaystyle M = 2e \approx 5.4366 }$

We are intended to find two solutions of equation $f(x)=\frac{ln(x)}{x}=r ,r<\frac{1}{e}$ . That is possible for all $x>1$ and not for any $x≤1$ . As the maximum value of $f(x)$ is $\frac{1}{e}$ .

Now, looking at the graph of $f(x)=\frac{ln(x)}{x}$ we see that the curve less steep on positive side of $x=e$ , the maximum point. Which means for some $x≠e$ , and $y$ ( the companion of $x$ , or $f(x)=f(y)$ $x<y$ ), $y-e>e-x \Rightarrow x+y>2e$ . Hence the $x$ for which $x+y$ is minimum is $e$ .

So, the answer is $2*e=5.437$ .

The curve of the function $f(x)=(x)^{\dfrac{1}{x}}$ is positively skewed. So, with growth of $x$ the average of two values of $x$ where $f(x)$ assumes the same value shifts towards right. Hence this average is minimum when $f(x)$ attains it's peak value of $(e)^{\dfrac{1}{e}}$ at $x=e$ . Therefore the minimum value of $x+y$ is $\boxed {2e}$