DSolve[{y''[x]==9/4 x (-y'[x])^(5/3),y[0]==1,y'[Infinity]==0 },y,x]

Test: solving equation (1) numerically

der0=-0.8059274488676563
(* Note that the exact shooting method does not apply because the solution we are seeking is an exceptional solution (sepatrix) *)
xlim=20
eqn=NDSolve[{y''[x]==9/4 x (-y'[x])^(5/3),y[0]==1,y'[0]==der0},y,{x,0,xlim}][[1]]
(* Note that we add the minus sign in (((y'))^(5/3)) to avoid the occurrence of complex numbers *)

(* Let us find the asymptotic solution. Invariance to the stretching group x->λ x, and y->λ^b y with b=-2. The asymptotic solution has the form y=α x^-2*)

Y[x_]:=a/x^2
Simplify[Y''[x]+9/4 x (Y'[x])^(5/3),Assumptions->{x>0}]

(* we use (-1)^5/3=1 and set α=4/3/(Sqrt[3])*)

α=(4/(3 (Sqrt[3])))
yas[x_]:=α x^-2

(* We plot the solution and check that it exhibits the expected asymptotic behaviour *)

des1=Show[Plot[y[x]/.eqn,{x,0,xlim},PlotRange->All],Plot[yas[x],{x,.75,xlim},PlotStyle->Dashed,PlotRange->All],PlotRange->All]

Method 1: using the translation group

(* The invariants are p=x and q=y'*)
(* y'' ==9/4 x (-y' )^(5/3) then becomes q'=9/4 p (-q)^(5/3) *)

(* A few steps of integration *)

Integrate[(3/4p^2+a)^(-3/2),p]

(* the solution is in the form when replacing p -> x *)

expr=b-((2 x)/(a (Sqrt[4 a+3 (x^2)])))

Limit[expr,x->Infinity]

(* let us find a and b such that y(0)=1 and y(∞)=0*)

(* value of b for which y(0)=1 : b=1*)

(* value of a for which y(0)=1 *)
Solve[-(2/((Sqrt[3]) a))+1==0,a]

Sol[x_,a_]:=2/a(1/Sqrt[3]-x(4a+3x^2)^(-1/2))

(* Plot *)
des2=Plot[Sol[x,(2/(Sqrt[3]))],{x,0,5},PlotRange->All,PlotStyle->Red]

(* Comparison with the numerical solution *)
Show[des1,des2]

(* Asymptotic behaviour *)
Series[-((2 x)/(a (Sqrt[4 a+3 (x^2)]))),{x,Infinity,3}]

Check

(* We check that the solution satisfies the ODE *)

4D[Sol[x,a],x,x]-9 x D[-Sol[x,a],x]^(5/3)
(* We need to help Mathematica to simply the equation *)
4D[Sol[x,a],x,x]-9 x D[-Sol[x,a],x]^(5/3)//FullSimplify[#,Assumptions->4 a+3 (x^2)>0]&

(* We check that the solution is invariant to the stretching group through an example by taking λ=2 *)

Show[Plot[Sol[x,1],{x,0,1}],Plot[Sol[2x,4] 4,{x,0,1},PlotStyle->{Dashed,Red}],PlotRange->All]

Method 2: using the stretching group

(* Plotting the phase portrait with the nullclines, the singular points O and A, and the tangent of the separatrix at A *)

pa=4/3/Sqrt[3];
qa=-2pa;
portrait=Show[StreamPlot[{1,q(3-9/4(Abs@q)^(2/3))/(2p+q)},{p,0,3},{q,0,-3}],
Plot[-2p,{p,0,3},PlotStyle->Black],Plot[-(4/3)^(3/2),{p,0,3},PlotStyle->Black],Plot[-4(p-pa)+qa,{p,0,3},PlotStyle->{Blue,Dashed}],Epilog->{
{PointSize[0.02],Red,Point[{4/3/Sqrt[3],-2*4/3/Sqrt[3]}]},
{PointSize[0.02],Red,Point[{0,0}]},
Text[Style["A",12,Bold],{4/3/Sqrt[3] ,-2*4/3/Sqrt[3]-0.25}],
Text[Style["O",12,Bold],{0,-0.25}]
}
]

(* Determining the slope of the separatrix at A *)

Solve[m==(3m-9/4 5/3 m (2 4/3/Sqrt[3])^(2/3))/(2+m),m]

(* Computing the sepatrix numericall and plotting it in the phase portrait *)
ϵ=10^(-3)
p0=pa-ϵ
sp=NDSolve[{q'[p]==q[p](3-9/4(Abs@q[p])^(2/3))/(2p+q[p]),q[p0]==qa+4ϵ},q,{p,p0,0}];
Show[portrait,Plot[q[p]/.sp,{p,p0,0},PlotStyle->Red]]

(* Returning to the original variables by numerical integration *)
sol[p_?NumberQ]:=q[p]/.sp[[1]]

neweq=NDSolve[{y'[x]==sol[x^2 y[x]]/x^3,y[ϵ]==1},y,{x,ϵ,10}]

des3=Plot[y[x]/.neweq,{x,ϵ,10},PlotStyle->Black]

Show[des1,des2,des3]