$$u_x+u^2u_y=1$$
I don't know exactly what you have done. So, I cannot comment it. It is certainly correct since your result is consistent with mine. I will only comment the manner to express and understand the result on my viewpoint.
For me, the general solution of the PDE is (on the form of implicit equation) :
$$\frac{1}{3}u^3-y=F(u-x)$$
where $F$ is any differentiable function.
$u(x,0)=1 \quad\implies\quad \frac{1}{3}1^3-0=F(1-x)=\frac{1}{3}$
This implies that the function $F$ is constant : $F(X)=\frac{1}{3}$ any $X$. So, the arbitrary function $F$ is now well determined according to the condition. Putting it into the general solution leads to :
$$\frac{1}{3}u^3-y=\frac13$$
$$u^3=3y+1$$
I cannot see any ambiguity up to now :
$\begin{cases} 3u^2u_y=3\quad\implies\quad u^2u_y=1\\ u_x=0\end{cases}\quad\implies\quad u_x+u^2u_y=0+1=1$
Thus, the solution $u^3=3y+1$ satisfies the PDE and the condition (even at $y=-\frac13$ ).
The apparent difficulty comes when you express the solution on the form :
$$u=\sqrt[3]{3y+1}$$
due to the derivative :$\quad u_y=\frac{1}{(3y+1)^{2/3}}\quad$ which is not finite at $y=-\frac13$.
But, in the PDE, the term $u^2u_y$ continue to be finite $=(\sqrt[3]{3y+1})^2\frac{1}{(3y+1)^{2/3}}=1$ at limit $y\to -\frac13$.
You can't have this:
$$\frac{du}{dx} = 3x/y ==> u (x,y) = 3x^2/y + f(C)$$
Because you have a differential equation ( du,dx) but three variables namely x,y,u. Y shouldnt be there...Apart for that mistake you did a very good job.
Usisng the method of characteristics:
$$\frac {dx}{y}=\frac {dy}{-x}=\frac {dz}{3x}$$
$$-xdx=ydy \implies x^2+y^2=K_1$$
$$3xdy=-xdz \implies y+z/3 =K_2$$
$$f(x^2+y^2)=y+u/3$$
$$u(x,y)=3f(x^2+y^2)-3y$$
we have $u(x,0)=x^2$
$$f(x^2)=x^2/3 \implies f(x)=x/3$$
Therefore
$$\boxed{u(x,y)=x^2+y^2-3y}$$
Best Answer
Observe that you equation can be written as $$ v_x=-v $$ where $v(x,y)=u_y-u$. Solving for $v$ (with $y$ as a parameter): $$ v(x,y)=C(y)e^{-x}. $$ where $C$ is an arbitrary function. So we have $$ u_y-u=C(y)e^{-x}. $$ which is a linear equation for $u$ with $x$ as a parameter. Solving you obtain $$ e^{-y}u=e^{-x}\int C(y)e^{-y} $$ or $$ u(x,y)=e^{y-x}\int C(y)e^{-y}\,dy. $$