Note that $y^2 y'^2 + \frac23 y^3$ is a constant of the motion, since its derivative is proportional to the equation of motion:

Simplify[D[y[x]^2 y'[x]^2 + 2/3 y[x]^3, x]] (* 2 y[x] y'[x] (y'[x]^2 + y[x] (1 + (y''[x])) *) 

(I found this by first noting that ${y'}^2 + y y'' = \frac{d}{dx}(y y') = \frac{1}{2} \frac{d^2 (y^2)}{dx^2}$ and then making further obvious substitutions from there.) Given the initial values $y(0) = 1$ and $y'(0) = 0$, we must have $y^2 y'^2 + \frac23 y^3 = \frac23$. Mathematica can solve this first-order ODE, though it outputs an InverseFunction involving a HyperGeometric2F1 function. As with @MichaelE2's answer, the resulting plot leaves something to be desired. Presumably the "real" solution you want would be the blue branch (which will be first function in the DSolve output.)

DSolve[{y[x]^2 y'[x]^2 + 2/3 y[x]^3 == 2/3, y[0] == 1}, y, x] Plot[Evaluate[y[x] /. %], {x, -2, 2}] 

Note that $y^2 y'^2 + \frac23 y^3$ is a first integral of this ODE, since its derivative is proportional to the ODE itself:

Simplify[D[y[x]^2 y'[x]^2 + 2/3 y[x]^3, x]] (* 2 y[x] y'[x] (y'[x]^2 + y[x] (1 + (y''[x])) *) 

(I found this by first noting that ${y'}^2 + y y'' = \frac{d}{dx}(y y') = \frac{1}{2} \frac{d^2 (y^2)}{dx^2}$ and then making further obvious substitutions from there.) Given the initial values $y(0) = 1$ and $y'(0) = 0$, we must have $y^2 y'^2 + \frac23 y^3 = \frac23$. Mathematica can solve this first-order ODE, though it outputs an InverseFunction involving a HyperGeometric2F1 function. As with @MichaelE2's answer, the resulting plot leaves something to be desired. Presumably the "real" solution you want for $x \in [0,1]$ would be the blue branch (which will be first function in the DSolve output.)

DSolve[{y[x]^2 y'[x]^2 + 2/3 y[x]^3 == 2/3, y[0] == 1}, y, x] Plot[Evaluate[y[x] /. %], {x, -2, 2}]