= Solution
For fixed $x>0$, set $y=y_0+\varepsilon y_1+\varepsilon^2y_2+\cdots$ and impose the data at one at each order. The leading equation gives $y_0'=0$, hence $y_0=1$. At the next two orders,
$$
x^2y_1'=1,\qquad x^2y_2'=2-\frac3x.
$$
The <boundary conditions> $y_1(1)=y_2(1)=0$ give the three-term <outer expansion>
$$
\boxed{y_{\rm out}=1+\varepsilon\left(1-\frac1x\right)
+\varepsilon^2\left(\frac12-\frac2x+\frac{3}{2x^2}\right)+\cdots}.
$$
Its ordering fails when $\varepsilon/x=O(1)$, locating a <boundary layer> of thickness $O(\varepsilon)$ at zero. Put $x=\varepsilon\xi$ and $y=Y_0(\xi)+\varepsilon Y_1(\xi)+\cdots$. The rescaled equation is
$$
(1+\varepsilon)\xi^2y_\xi=y^3+
\varepsilon[\xi(y^2-1)+2y^2]-\varepsilon^2\xi(y^2+1).
$$
Its leading equation is $\xi^2Y_0'=Y_0^3$. Matching to the positive outer value one fixes the sign and integration constant:
$$
Y_0=\left(\frac\xi{\xi+2}\right)^{1/2}.
$$
At the next order the bracket $\xi(Y_0^2-1)+2Y_0^2$ vanishes identically, leaving
$$
\xi^2Y_1'-\frac{3\xi}{\xi+2}Y_1
=-\left(\frac\xi{\xi+2}\right)^{3/2}.
$$
The general solution is $Y_1=(1+k\xi)\xi^{1/2}/(\xi+2)^{3/2}$. In the overlap its expansion is
$$
Y_0=1-\frac1\xi+\frac{3}{2\xi^2}+\cdots,\qquad
Y_1=k+\frac{1-3k}{\xi}+\cdots.
$$
The outer solution re-expressed on this scale has the corresponding terms $1-1/\xi+3/(2\xi^2)+\varepsilon(1-2/\xi)+\cdots$. Thus <matched asymptotic expansion> determines $k=1$, consistently in both displayed orders, and the two-term <inner expansion> is
$$
\boxed{y_{\rm in}(\xi)=\left(\frac\xi{\xi+2}\right)^{1/2}
+\varepsilon\frac{(1+\xi)\xi^{1/2}}{(\xi+2)^{3/2}}+\cdots}.
$$
This positive branch tends continuously to zero at $x=0$, with a square-root cusp. An infinite endpoint <derivative> is compatible with the degeneracy of the original equation's $x^2y'$ coefficient.
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