Consider the following general quasilinear equation, $u(x,y): \mathbb{R}^2 \to \mathbb{R}$:
$$ A(u,x,y) u_x + B(u,x,y) u_y + C(u,x,y) = 0, (x,y) \in \Omega \subset \mathbb{R}^2 $$
with respect to the boundary data $u(0,y) = h(0,y), y \in \Gamma \subset \partial \Omega$.
I was told that one could recover all partial derivatives of $u$ from $h$ at the boundary provided the non-characteristic condition holds, namely: $$ (A(u,0, y_0), B(u,0,y_0)) \cdot (\vec{n}_1, \vec{n}_2) \neq 0 $$
where $\vec{n}_1, \vec{n}_2$ denotes the outward unit normal of $\partial \Omega$. I don't see why.
Since the boundary is specified at the $y$-axis $x = 0$, the unit normal vector is horizontal and so your non-characteristic condition simply reads $A(u,0,y) \neq 0$. Now since $u(0,y)$ is specified, clearly we can recover $u_y$ on the boundary. But that means we can calculate the $u_x$ partial since by the PDE $$ A(u,0,y)u_x + B(u,0,y)u_y + C(u,0,y) = 0$$ for every $y$ and so $$ \implies u_x = -\frac{B(u,0,y)u_y + C(u,0,y)}{A(u,0,y)}$$ on the boundary. Notice that the non-characteristic condition was essential here.
Edit: I'll give a concrete example. If we are trying to solve $uu_x + (x + y)u_y = \cos(xy)$ with the boundary condition $u(0,y) = e^{2y}$ for every $y$, then we easily calculate $u_y(0,y) = 2e^{2y}$ for every $y$. Furthermore, since the non-characteristic condition $u = e^{2y} \neq 0$ holds for every $y$, we can also calculate from the PDE $$e^{2y}u_x(0,y) + y(2e^{2y}) = 1 \implies u_x(0,y) = \frac{1-2ye^{2y}}{e^{2y}}$$ along the boundary. Therefore both partials are known for $x = 0$.