= Solution
For $\eta=x/y^\beta$ the <chain rule> gives $u_y=-\beta\eta f'/y$, $u_x=y^{-\beta}f'$ and $(uu_x)_x=y^{-2\beta}[ff''+(f')^2]$. Multiplication of the equation by $y$ gives
$$
-\beta\eta f'=y^{1-2\beta}[ff''+(f')^2]+\tfrac12f.
$$
A nontrivial scale-invariant balance therefore selects $\boxed{\beta=1/2}$ and yields
$$
ff''+(f')^2+\tfrac12\eta f'+\tfrac12f=0.
$$
Its left side is the <derivative> of $ff'+\eta f/2$. The <boundary conditions> give $f(0)=1$ and $f'(0)=0$, so the integration constant is zero:
$$
f(f'+\eta/2)=0.
$$
On the positive branch before the first zero, divide by $f$ and integrate, obtaining $f=1-\eta^2/4$. Its first positive zero is $\eta_F=2$. Hence
$$
\boxed{u(x,y)=1-\frac{x^2}{4y}\quad(0\leq x\leq2\sqrt y),\qquad x_F(y)=2\sqrt y.}
$$
A direct check gives $u_y=x^2/(4y^2)$ and $(uu_x)_x+u/(2y)=x^2/(4y^2)$. The zero extension past this front is also a natural weak compact-support continuation: $u$ is continuous and the degenerate diffusive <flux> $uu_x$ tends to zero at the front. This is a <source-driven porous-medium similarity front>.
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