= Solution
Take an admissible variation $y+\varepsilon\eta$ with $\eta=\eta'=0$ at both regular endpoints. Its <first variation> is $\delta I=\int(f_y\eta+f_{y'}\eta'+f_{y''}\eta'')\,dx$. Integrating the second term once and the third twice gives the boundary term $[f_{y'}\eta+f_{y''}\eta'-(d f_{y''}/dx)\eta]_a^b$, which vanishes. The <fundamental lemma of the calculus of variations> therefore yields the <higher-order Euler-Lagrange equation>
$$
\boxed{f_y-\frac d{dx}f_{y'}+\frac{d^2}{dx^2}f_{y''}=0.}
$$
Here $f_y=8yy'$, $f_{y'}=4y^2$, and $f_{y''}=2x^4y''$. The first two contributions cancel, leaving $(x^4y'')''=0$. For $x>0$, integrate twice to obtain $x^4y''=Ax+B$, and then
$$
y(x)=\frac A{2x}+\frac B{6x^2}+Cx+D.
$$
The two singular-endpoint limits give $B=0$ and $A=2$. The conditions at one give $C=1$ and $D=0$. Hence the unique solution of the stated differential equation and limiting data is
$$
\boxed{y(x)=x+\frac1x\qquad(0<x\le1).}
$$
There is a genuine distinction between solving this equation and stationarity of a finite improper action. Substitution gives $f=4x^2+4-4x^{-4}$, whose <integral> at zero diverges to $-\infty$. The displayed Euler-Lagrange equation remains valid for compactly supported variations on $(0,1)$, or on regular cut-off intervals, but the literal unregularized action is not a finite stationary functional on this solution. These <singular endpoints in higher-order variational problems> require a regularization or an admissible-space convention if a global variational claim is intended.
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