I have solved
$$yu_x+u_y=u^2$$
using $x=-\frac{s^2}{2},\,y=s>0,\,u=1$ and obtained the solution
$$u(x,y)=\frac{1}{\sqrt{\frac{y^2}{2}-x}+1-y}$$
Now I am trying to show $\textbf{A}\cdot\nabla G=0$ where
$$G(x,y,u)=\sqrt{\frac{y^2}{2}-x}+1-y-\dfrac{1}{u}=0, \qquad\textbf{A}=(y,1,u^2)$$
Therefore, as
$$\nabla G = \left(-\frac{1}{2}\left(\frac{y^2}{2}-x\right)^{-\frac{1}{2}}+\dfrac{u_x}{u^2},\,\frac{y}{2}\left(\frac{y^2}{2}-x\right)^{-\frac{1}{2}}-1+\dfrac{u_y}{u^2},\,\frac{1}{u^2}\right)$$
I find that
\begin{align} \textbf{A}\cdot{\nabla}G &= -\frac{y}{2}\left(\frac{y^2}{2}-x\right)^{-\frac{1}{2}}+y\dfrac{u_x}{u^2}+\frac{y}{2}\left(\frac{y^2}{2}-x\right)^{-\frac{1}{2}}-1+\dfrac{u_y}{u^2}+1 \\ &= y\frac{u_x}{u^2}+\frac{u_y}{u^2} \\ &= 1 \end{align}
This should be equal to zero though so where have I gone wrong?
The gradient of $G(x,y,u)$ (note that $u$ is a free variable) should be $$\nabla G(x,y,u)=\left(-\frac{1}{2}\left(\frac{y^2}{2}-x\right)^{-\frac{1}{2}},\,\frac{y}{2}\left(\frac{y^2}{2}-x\right)^{-\frac{1}{2}}-1,\,\frac{1}{u^2}\right)$$ and then we find $$\textbf{A}\cdot{\nabla}G=-\frac{y}{2}\left(\frac{y^2}{2}-x\right)^{-\frac{1}{2}}+\frac{y}{2}\left(\frac{y^2}{2}-x\right)^{-\frac{1}{2}}-1+1=0.$$ I try to put it in other words. Let $u=h(x,y)$ be any solution of $yu_x+u_y=u^2$ and define $G(x,y,u)=\frac{1}{h(x,y)}-\frac{1}{u}$. Then $$\nabla G(x,y,u)=\left(-\frac{h_x}{h^2},-\frac{h_y}{h^2},\frac{1}{u^2}\right)$$ and $$\textbf{A}\cdot{\nabla}G=(y,1,u^2)\cdot \left(-\frac{h_x}{h^2},-\frac{h_y}{h^2},\frac{1}{u^2}\right)=-\frac{yh_x}{h^2}-\frac{h_y}{h^2}+1=0.$$ Therefore the verification of $\textbf{A}\cdot{\nabla}G=0$ can be done without solving $yu_x+u_y=u^2$ explicitly.