I would like to find the leading behavior at the irregular singular point at $x=0$.
First substitute $y = e^{S(x)}$, we obtain $$S'' + (S')^2 = \sqrt x \quad \quad (\star)$$ testing the two assumptions that $S'' << (S')^2$ and $(S')^2 << S''$ as $x\rightarrow 0$, we see that $S''$ is the dominating term which gives $$S'' \sim \sqrt x,$$ so $S\sim \frac{4}{15}x^{5/2}$ as $x\rightarrow 0$.
Now assume $S(x) = \frac{4}{15}x^{5/2} + C(x)$, from $S'' \sim \sqrt x$, I think we get $C'' << \sqrt x$ and $C' << x^{3/2}$. Plug this new $S$ into $(\star)$, we get $$C'' + \frac{4}{9} x^3 + 2\frac{2}{3}x^{3/2} C' + (C')^2 = 0$$ I know $(C')^2 << 2\frac{2}{3}x^{3/2} C'$ so $$C'' + \frac{4}{9} x^3 + 2\frac{2}{3}x^{3/2} C'\sim 0 $$ but how could I simplify it further to something that can easily solve for $C$?
Try $y(x)=x^rf(x^s)=\sum_{k=0}^\infty a_kx^{r+sk}$, then $$ \sum_{k=0}^\infty a_k(r+sk)(r+sk-1)x^{r+sk-2}=y''(x)=\sqrt{x}y(x)=\sum_{k=0}^\infty a_kx^{r+sk+\frac12}. $$ Identifying the second term on the left with the first term on the right and so on gives $r+(k+1)s-2=r+ks+\frac12\implies s=\frac52$. The indicial equation $r(r-1)=0$ has $r=0$ or $r=1$. The further coefficients follow the recursion $$ a_{k}=\frac1{(r+\frac52k)(r+\frac52k-1)}a_{k-1} $$ so that the first coefficients are (with $a_0=1$) \begin{array}{cc|c}&r=0&r=1\\\hline a_1=&\frac{4}{15}&\frac4{35}\\ a_2=&\frac1{75}&\frac2{525}\\ ... \end{array} or inserted into the expansion of the solutions $$ y_1(x)=1+\frac{4}{15}x^{\frac52}+\frac1{75}x^5+...\\ y_2(x)=x\left(1+\frac4{35}x^{\frac52}+\frac2{525}x^5+...\right) $$