Solve $x^4y^{\prime\prime} = (y-xy^\prime)^3, y(1) = y^\prime(1) = 1$ $\lambda^4x^4\lambda^{n-2}y^{\prime\prime} = (\lambda^ny-\lambda x\lambda^{n-1}y^\prime)^3 \Rightarrow \lambda^{n+2}x^4y^{\prime\prime} = (\lambda^n(y-xy^\prime))^3$. 
$\lambda^{n+2} = \lambda^{3n} \Rightarrow n+2 = 3n \Rightarrow n = 1$.
Let $x = e^t, y = ue^{nt} = ue^t$.
Now, $\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{u^\prime e^t + ue^t}{e^t} = u^\prime + u$. And, $\frac{d^2y}{dx^2} = \frac{\frac{d}{dt}}{\frac{dx}{dt}}(\frac{dy}{dx}) = \frac{\frac{d}{dt}(u^\prime + u)}{e^t} = e^{-t}(\frac{d^2u}{dt^2} + \frac{du}{dt})$.
Thus,
$e^{4t}(e^{-t}(\frac{du^2}{dt^2} + \frac{du}{dt})) = (ue^{t}-e^t(u+\frac{du}{dt}))^3 \Rightarrow e^{3t}(\frac{d^2u}{t^2} + \frac{du}{dt}) = e^{3t}(u-(u+\frac{du}{dt}))^3 \Rightarrow (\frac{d^2u}{dt^2} + \frac{du}{dt}) = (\frac{du}{dt})^3$.
Let $p = \frac{du}{dt}, p^\prime = \frac{d^2u}{dt^2} = \frac{dp}{du}\frac{du}{dt} = p\frac{dp}{du}$.
Thus,
$(p + p\frac{dp}{du}) = p^3 \Rightarrow 1 + \frac{dp}{du} = p^2 \Rightarrow \frac{dp}{du} = p^2-1 \Rightarrow u + c_1 = \int\frac{dp}{p^2-1} = \int\frac{dp}{(p-1)(p+1)} = \int(\frac{1}{2(p-1)} - \frac{1}{2(p+1)})dp = \frac{1}{2}\ln(p-1)-\frac{1}{2}\ln(p+1) = \ln(\frac{\sqrt{p-1}}{\sqrt{p+1}}) \Rightarrow c_1e^u = \frac{\sqrt{p-1}}{\sqrt{p+1}} \Rightarrow c_1e^{2u} = \frac{p-1}{p+1} \Rightarrow p-1 = c_1e^{2u}p + c_1e^{2u} \Rightarrow p = \frac{c_1e^{2u}+1}{1-c_1e^{2u}} = \frac{du}{dt}$.
But then I think that I have to use the initial conditions to get $c_1$, but I don know how to do that.

Since $x=e^t$, $x=1$ exactly when $t=0$. So, since $y=ue^{t}$, the boundary condition $y(1)=1$ translates into $u(0)e^0=1$ which means $u(0)=1$.
Then use the fact that $y'=u'+u$. The boundary condition $y'(1)=1$ means $u'(0)+u(0)=1$. Since we already have that $u(0)=1$, this means $u'(0)=0$.
You have already partially solved for $u'$, because you have $u'=\frac{c_1e^{2u}+1}{1-c_1e^{2u}}$. This gives you that $u'(0)=\frac{c_1e^{2u(0)}+1}{1-c_1e^{2u(0)}}=\frac{c_1e^2+1}{1-c_1e^2}$. Setting this last expression equal to $0$, you get $c_1=-\frac{1}{e^2}$.  
(Keep in mind that terms like $y'$ and $u'$ implicitly denote derivatives with respect to $x$ and $t$, respectively).
