How to solve this equation iteratively?

Something like the following? Here is your equation:

eqn =  x^2 == (Subscript[a, 4] K^6)/(9 x^6)+(Subscript[a, 3] K^4)/(7x^4) +
    (Subscript[a, 2] K^2)/(5 x^2)+Subscript[a, 1]/3;
eqn //TeXForm

$x^2=\frac{a_4 K^6}{9 x^6}+\frac{a_3 K^4}{7 x^4}+\frac{a_2 K^2}{5 x^2}+\frac{a_1}{3}$

Construct a rule converting x to a series in K (still including x):

rule = x -> Sqrt[Series[eqn[[2]], {K, 0, 8}]];
rule //TeXForm

$x\to \frac{\sqrt{a_1}}{\sqrt{3}}+\frac{\sqrt{3} a_2 K^2}{10 x^2 \sqrt{a_1}}+\frac{\sqrt{a_1} \left(\frac{6 a_3}{7 x^4 a_1}-\frac{9 a_2^2}{50 x^4 a_1^2}\right) K^4}{4 \sqrt{3}}+\frac{\sqrt{a_1} \left(\frac{a_4}{x^6 a_1}-\frac{9 a_2 \left(\frac{6 a_3}{7 x^4 a_1}-\frac{9 a_2^2}{50 x^4 a_1^2}\right)}{20 x^2 a_1}\right) K^6}{6 \sqrt{3}}+\frac{\sqrt{a_1} \left(-\frac{3 a_3 \left(\frac{6 a_3}{7 x^4 a_1}-\frac{9 a_2^2}{50 x^4 a_1^2}\right)}{14 x^4 a_1}+\frac{a_2 a_4}{10 x^8 a_1^2}-\frac{a_2 \left(\frac{a_4}{x^6 a_1}-\frac{9 a_2 \left(\frac{6 a_3}{7 x^4 a_1}-\frac{9 a_2^2}{50 x^4 a_1^2}\right)}{20 x^2 a_1}\right)}{2 x^2 a_1}\right) K^8}{8 \sqrt{3}}+O\left(K^9\right)$

Use this rule repeatedly until you get the result at the desired order (and free of x):

x^2 //. rule //TeXForm

$\frac{a_1}{3}+\frac{3 a_2 K^2}{5 a_1}+\frac{9 \left(25 a_1 a_3-21 a_2^2\right) K^4}{175 a_1^3}+\frac{3 \left(1134 a_2^3-2025 a_1 a_3 a_2+875 a_1^2 a_4\right) K^6}{875 a_1^5}-\frac{27 \left(3969 a_2^4-9450 a_1 a_3 a_2^2+4900 a_1^2 a_4 a_2+2250 a_1^2 a_3^2\right) K^8}{6125 a_1^7}+O\left(K^9\right)$

Here is a little different approach that yields the same result:

eqn = x^2 == (Subscript[a, 4] K^6)/(9 x^6) + (Subscript[a, 
        3] K^4)/(7 x^4) + (Subscript[a, 2] K^2)/(5 x^2) + 
    Subscript[a, 1]/3 /. x -> Sqrt[x2]

First sub a series expression for x2 (=x^2) on both sides of the equation, then take Series to the whole thing , finally require all the K coefficients to be zero..

vars = Table[b[i], {i, 0, 8}]
sol = Solve[
  CoefficientList[
    Normal@Series[ 
      eqn[[1]] - eqn[[2]] /. x2 -> Sum[b[i] K^i , {i, 0, 8}] , {K, 0, 
       8}], K] == 0, vars]

 Sum[b[i] K^i , {i, 0, 8}] /. sol

same result

You can also use the new in M12 function AsymptoticSolve. First, it will be simpler to use a new variable x2 = x^2:

eqn = x2 == (Subscript[a, 4] K^6)/(9 x2^3) + (Subscript[a, 3] K^4)/(7x2^2) +
    (Subscript[a, 2] K^2)/(5 x2) + Subscript[a, 1]/3;
eqn //TeXForm

$\text{x2}=\frac{a_4 K^6}{9 \text{x2}^3}+\frac{a_3 K^4}{7 \text{x2}^2}+\frac{a_2 K^2}{5 \text{x2}}+\frac{a_1}{3}$

At K=0, x^2 is given by $a_1/3$, so the call to AsymptoticSolve is:

x2 /. First @ AsymptoticSolve[eqn, {x2, Subscript[a,1]/3}, {K, 0, 6}] //TeXForm

$\frac{3 \left(1134 a_2^3-2025 a_1 a_3 a_2+875 a_1^2 a_4\right) K^6}{875 a_1^5}+\frac{9 \left(25 a_1 a_3-21 a_2^2\right) K^4}{175 a_1^3}+\frac{3 a_2 K^2}{5 a_1}+\frac{a_1}{3}$

in agreement with my other answer.

Before you have access to M12, you can use CloudEvaluate:

$VersionNumber
x2 /. CloudEvaluate @ First @ System`AsymptoticSolve[
    eqn,
    {x2, Subscript[a,1]/3},
    {K, 0, 6}
] //TeXForm

11.3

$\frac{3 \left(1134 a_2^3-2025 a_1 a_3 a_2+875 a_1^2 a_4\right) K^6}{875 a_1^5}+\frac{9 \left(25 a_1 a_3-21 a_2^2\right) K^4}{175 a_1^3}+\frac{3 a_2 K^2}{5 a_1}+\frac{a_1}{3}$