Find the vector field associated to the PDE and define a parametrization of the curve that define a border condition $$(u+2y)u_{x}+uu_{y}=0$$ with $u(x,1)=\frac{1}{x}$.
My approach: Consider the following first-order, linear equation $$a(x,y)u_{x}+b(x,y)u_{y}=c(x,y)$$ Suppose we can find a solution $u(x,y)$. let’s start by constructing a curve $C$ parametrized by $s$ such that at each point on the curve $C$, the vector $(a(x(s), y(s)), b(x(s), y(s)), c(x(s), y(s)))$ is tangent to the curve. In particular, $$\frac{dx}{ds}=a(x(s),y(s))$$ $$\frac{dy}{ds}=b(x(s),y(s))$$ $$\frac{dz}{ds}=c(x(s),y(s))$$
Which are the integral curve for $(a(x, y), b(x, y), c(x, y))$. So, the vector field is given by $V=(a(x,y),b(x,y),c(x,y))$? In this case, we have $V=(u+2y,u,0)$?
And how define the parametrized curve that define the boundary condition?
$$(u+2y)u_{x}+uu_{y}=0\tag 1$$ I agree with $$V=(u+2y,u,0)$$ Thus your system of differential equations is $$\frac{dx}{ds}=a(x(s),y(s))=u+2y$$ $$\frac{dy}{ds}=b(x(s),y(s))=u$$ $$\frac{du}{ds}=c(x(s),y(s))=0$$ a short writing is : $$\frac{dx}{u+2y}=\frac{dy}{u}=\frac{du}{0}=ds$$ A first characteristic curve comes from $du=0$ since $u$ is finite. Thus $u=c_1$.
A second characteristic curve comes from $\frac{dx}{c_1+2y}=\frac{dy}{c_1}$ which integrating leads to : $$c_1y+y^2-c_1x=c_2$$ The general solution of the PDE $(1)$ expressed on the form of implicit equation is : $$\Phi\left(c_1,c_2 \right)=\Phi\left(u\:,\:uy+y^2-ux \right)=0$$ where $\Phi$ is any function of two variables.$
Or equivalently : $$u=F\left(uy+y^2-ux \right) \tag 2$$ where $F$ is an arbitrary function.
BOUNDARY CONDITION :
$u(x,1)=\frac{1}{x}=F\left(\frac{1}{x}+1^2-\frac{1}{x}x \right)$. $$\frac{1}{x}=F\left(\frac{1}{x} \right)$$. So, the function $F$ is determined : $\quad F(X)=X$
Putting this function $F(X)=X$ into the general solution $(2)$ where $X=uy+y^2-ux $ leads to $$u=uy+y^2-ux $$ Then, solving it for $u$ : $$u=\frac{y^2}{1-y+x}$$ which is the particular solution of the PDE fitting to the boundary condition.