**Analysis of the equation**

It often is convenient to non-dimensionalize ODEs arising in physics, and this case is no exception.  Rescale `x` and `y` by `{x -> xn xs, y[x] -> yn[xn] ys}`.  Apply these transformations to the ODE, bringing the scale factors from the left side of the ODE. (xs^2/ys)`, to the right.

    eqlast = (xs^2/ys) k ((e y[x]/c)^2 + 2 m*e y[x])^(3/2) /. 
        y[x] -> yn[t] ys /. x -> xn xs /. t -> xn
    (* (k xs^2 (2 e m ys yn[xn] + (e^2 ys^2 yn[xn]^2)/c^2)^(3/2))/ys *)

Now, choose the scale factors, `xs` and `ys`, to eliminate the multipliers of `yn[xn]` and `yn[xn]^2`.

    eq1 = eqlast[[1 ;; 3]] (eqlast[[4, 1, 1]])^(3/2) /. yn[xn] -> 1 /. xn -> 1
    (* (2 Sqrt[2] k xs^2 (e m ys)^(3/2))/ys *)
    eq2 = eqlast[[1 ;; 3]] (eqlast[[4, 1, 2]])^(3/2) /. yn[xn] -> 1 /. xn -> 1
    (* (k xs^2 ((e^2 ys^2)/c^2)^(3/2))/ys *)
    norms = FullSimplify[Solve[eq1 == 1 && eq2 == 1 && k > 0 && e > 0 && m > 0 && c > 0, 
        {xs, ys}, Reals], k > 0 && e > 0 && m > 0 && c > 0] // Last
    (* {xs -> Sqrt[1/(c e k)]/(2 m), ys -> (2 c^2 m)/e} *)

Inserting these expressions into the transformed ODE yields,

    eq = 1/xn D[xn yn[xn], {xn, 2}] == Simplify[eqlast /. %, k > 0 && e > 0 && m > 0 && c > 0]
    (* (2 yn'[xn] + xn yn''[xn])/xn == (yn[xn] (1 + yn[xn]))^(3/2) *)

which is somewhat easier to work with than the ODE in its original form.  The question seeks a solution with very small `y` at large `x`.  To obtain such an asymptotic solution, assume that `y^2` is much smaller than `y` (in absolute value).  Then, it is apparent that a solution exists equal to `cf xn^a`, `a` and `cf` constants.

    Unevaluated[1/xn D[xn yn[xn], {xn, 2}] - (yn[xn] (1 + yn[xn]))^(3/2)] /. 
        {(1 + yn[xn]) -> 1, yn[xn] -> cf xn^a}
    (* a (1 + a) cf xn^(-2 + a) - (cf xn^a)^(3/2) *)

For this expression to vanish, xn^(-2 + a) == xn^(3 a/2) clearly is required.  In other words, `a == -4`.  Then, `cf` is determined from

    Simplify[% /. a -> -4, cf > 0 && xn > 0]
    (* -(((-12 + Sqrt[cf]) cf)/xn^6) *)

So, `cf == 144`, and this particular solution is `yn[xn] == 144/xn^4`.  As we shall see, this is not the most general solution at large `xn`, but it does appear to be a separatrix.

**Numerical Solution`**

With the constants given in the question, the scale factors become,

    norms
    (* {xs -> 3.46944*10^-12, ys -> 1.022*10^6} *)

and the starting point of the integration, `x0`, normalized to `xs`, becomes (rationalized for future use in `NDSolve`)

    x0n = Rationalize[x0/xs /. norms, 0]
    (* 723020393/5483051 *)

or about `131.865`, and the specified value of `y[x0]`, normalized to `ys`, becomes

    y0n = Rationalize[10^6/ys /. norms, 0]
    (* 33547031/34284995 *)

or about `0.978476`.  Conveniently, `x0n` is well away from the singularity at `xn == 0`.  The problem, now, is to choose `yn'[x0n]` to that `yn[xn]` approaches the separatrix, obtained above, at large `xn`.  Employing a variant on the method employed to solve question [147207][1] yields the desired value of `yn'[x0n]`.

    xfn = 1500;
    s = ParametricNDSolve[{eq, yn[x0n] == y0n, yn'[x0n] == yp0, 
        WhenEvent[Re@yn[xn] < 120/xfn^4 || yn'[xn] > 1/100, "StopIntegration"]}, 
        yn, {xn, x0n, xfn}, {yp0, wp}, WorkingPrecision -> wp, AccuracyGoal -> Infinity, 
        MaxSteps -> 20000];

`WhenEvent[ ... ]` is used here to terminate integration, whenever a numerical solution veers away from the separatrix.  Doing so is desirable, because one hundred or more iterations at high `WorkingPrecision` typically are needed to obtain a desired solution.  A small amount of experimentation is sufficient to determine that the initial value for `yn'[x0n]` is near `-1.39`.  A better approximation then is obtained from

    plt = Plot[Quiet@(yn[yp0, 15]["Domain"] /. s)[[1, 2]], {yp0, -1.306, -1.304}, 
        PlotRange -> All]

[![enter image description here][2]][2]

The improved estimate, given by the spike in the plot, is located within the range,

    Cases[plt, Line[a__] :> a, Infinity] // Last;
    Position[%, Max[Last /@ %]][[1, 1]];
    est = SetPrecision[{First[%%[[% - 1]]], First[%%[[% + 1]]]}, 45]
    (* {-1.30493508434085847547123648837441578507423401, 
        -1.30493385315771526222761167446151375770568848} *)

This already narrow range is reduced much further with

    dom[yp0_?NumericQ] := Quiet@(yn[yp0, 45]["Domain"] /. s)[[1, 2]] 
    gg[bl0_, bu0_] := Module[{bl = N[bl0, 45], bu = N[bu0, 45], bt, db}, 
        Do[bt = (bl + bu)/2; db = (bu - bl)/100; 
        If[dom[bt] - dom[bt - db] > 0, bl = bt, bu = bt], {i, 120}]; bt];

(Note that `FixedPoint` could be used here instead of `Do` but appears to offer no advantage, at least for the present computation.)  

    rough = gg @@ est
    (yn[rough, 45][xfn] /. s) - 120/xfn^4
    (* -1.30493476864289837272479116355841854529571009 *)
    (* 2.170763397571277507505428*10^-30 *)

About `80 sec` was used by my PC to obtain this result.  Such very high precision usually is necessary when seeking a solution asymptotically approaching a separatrix.  Below is a plot of the numerical result, in blue, and the separatrix, in orange.

    LogPlot[{yn[rough, 45][xn] /. s, 144/xn^4}, {xn, x0n, xfn}, 
        AxesLabel -> {"x/xs", "y/ys"}, LabelStyle -> Directive[Bold, 11], PlotRange -> All]

As expected (and desired), the numerical solution is quite near the separatrix at large `xn`, even when viewed on a log plot.

[![enter image description here][3]][3]


  [1]: https://mathematica.stackexchange.com/a/147585/1063
  [2]: https://i.sstatic.net/BkEwK.png
  [3]: https://i.sstatic.net/D33M4.png