I was trying to solve this probleme using the method of differentials :
$\begin{cases} \frac{\partial u}{\partial x} +u\frac{\partial u}{\partial y} = 0 ~~~~~~, (x,y)\in \mathbb{R}^{*}_{+} \\u(x,0)= x ~~~~, \end{cases}$
where $u=u(x,y)$ , since we have :
$\begin{cases} \frac{dx}{1}=\frac{dy}{u} \\ du = 0 \end {cases}$
then : $~~~~ u(x,y)= B$ $~~~~$and $~~~x=\frac{y}{B} +A$
setting : $~~B=f(A)$ $~~~~$ we get : $~~u(x,y)=f(x-\frac{y}{B}) = f(x-\frac{y}{u})$
using the initial condition condition : $~~u(x,0)=f(x) =x$
which means : $~~u(x,y)= x-\frac{y}{u}$ .
This is the best form I managed to get , is there any other alternative which could possibly lead to an explicit form of $u$ ?
I agree with your general solution : $\quad u=f(x-\frac{y}{u})$
The condition $u(x,0)=x$ determines $f(x)=x$. So, your particular solution fitting the condition is correct (on the form of an implicit equation) : $$u=x-\frac{y}{u}$$ In order to express it on the explicit form, you have to solve it for $u$:
$u^2-xu+y=0$
$u=\frac12(x\pm\sqrt{x^2+4y})$
With the condition and on $(x,y)\in \mathbb{R}^{*}_{+}$ the sign of the square root is determined : $$u(x,y)=\frac12\left(x+\sqrt{x^2+4y}\right)$$