Given the IVP $$ y' = \sqrt{1-y^2},\qquad y(0) = 0 \tag{1}$$ I'm looking for a solution in the form of a power series in power of $x$ about $x=0$. Particularly, I'm looking for the coefficients up to the term in $x^{3}$ in the series.
My attempt is as follows ; the given ODE $(1)$ is separable hence \begin{align} \int \frac{dy}{\sqrt{1-y^2}} &= \int dx \\ \sin^{-1}(y) &= x+C, \end{align} given the initial condition $y(0)=0$, \begin{gather} y = \sin(x+0) = \sin(x)\\ \therefore y(x) = \sum_{k\ge 0} \frac{(-1)^k x^{2k+1}}{(2k+1)!} = x - \frac{x^3}{3!} + \dots. \end{gather} Therefore, the coefficient sup to $x^3$ as requested are $\{1, -\frac13 \}$. I'm confused by two things : the simplicity of the reasoning and I'm not sure if the first coefficient is actually $0$ instead of $1$ , since the initial condition dictates the value of the initial term in a series solution (I think).
Of course the solution $y(x)=\sin(x)$ is obvious by inspection or by integration as you did. Note that the coefficient of $x^3$ is not $-\frac13$ but is $-\frac16$. A typo I suppose.
Are you sure that this is what is asked for ?
The question might be to find the coefficient of $x^3$ without solving the ODE, but only with series expansion :
$$y(x)=a_1x+a_2x^2+a_3x^3+O(x^4) \quad ;\quad a_0=0 \text{ because}\quad y(0)=0.$$
$$y'=a_1+2a_2x+3a_3x^2+O(x^3)$$ $$y'=a_1+2a_2x+3a_3x^2+O(x^3)=\sqrt{1-\big(a_1x+a_2x^2+O(x^3)\big)^2}$$ $$a_1+2a_2x+3a_3x^2+O(x^3)=1-\frac12\big(a_1x+O(x^2)\big)^2=1-\frac12 a_1^2x^2+O(x^3)$$ $$(a_1-1)+2a_2x+(3a_3+\frac12 a_1^2)x^2+O(x^3)=0\quad\implies\quad \begin{cases} a_1=1 \\ a_2=0 \\ a_3=-\frac16 \end{cases} $$