I got a weird result so I'm not sure I did this right
Let the initial condition be $u(0, y_0) = y_0 $ for some $y_0$
By the method of characteristics let
$$\frac{dx}{ds} = 1 \to x = s + A$$ $$x(s=0) = A = 0 \to x(s) = s$$
$$\frac{dy}{ds} = 4x = 4s \to y = 2s^2 + B$$ $$y(s=0) = B = y_0 \to y(s) = 2s^2+y_0$$
$$\frac{du}{ds} = 1 + u^2 \to \arctan u = s + C \to u(s)=\tan(s+C)$$ $$u(0) = \tan(C) = y_0 \to C=\arctan y_0$$
Now we have that $$u(s) = \tan(s+C) = \tan(x + \arctan y_0)$$
Using $$y(s) = 2s^2+y_0 \to y_0 = y - 2s^2$$
We can substitute in $u$ to get $$u(s) = \tan(x + \arctan(y-2s^2)) = \tan(x + \arctan(y-2x^2)) = u(x,y)$$