I'm assuming you know the method of characteristics, so we want the solution $\xi$ to be constant in the direction $(1,-p_1)$ (obtain by looking at the coefficients of $\partial_x\xi$ and $\partial_y\xi$). So we want the derivative in the direction $(1,-p_1)$ to be zero, i.e.,
$$\nabla \xi\cdot(1,-p_1)=0.$$
(remember $\nabla\xi\cdot(1,-p_1)=(\partial_x\xi,\partial_y\xi)\cdot(1,-p_1)=\partial_x\xi-p_1\partial_y\xi$).
This holds for $\xi(x,y) = x-p_1y$.
We have that $\Delta=0$, this is the discriminant of the quadratic equation $4x^2-12x+9x^2=0$, which thus has a repeated root $x=\frac{3}{2}$, so we can express $$4\partial^2_{xx}u-12\partial^2_{xy}u+9\partial^2_{yy}u=4(\partial_x -\frac{3}{2}\partial_y)^2u.$$
Dividing our original equation by $4$ we obtain:
$$(\partial_x -\frac{3}{2}\partial_y)^2u+\frac{1}{4}\partial_xu=0,$$
so $p_1=\frac{3}{2}.$
I have now solved it by putting $x^3 = (\xi - \eta)^{3/2}$ and then integrating the canonical equation will give the given solution.
Edit(Complete solution):
Comparing the given PDE with the general form $A(x, y)u_{xx} + B(x, y)u_{xy} + C(x, y)u_{yy} = Φ(x, y,u,u_x,u_y)$ we find
$$A = x, B= 2x^2, C = 0$$
Now, the characteristic curves can be found by the equation $\frac{dy}{dx} = \frac{B \pm\sqrt{B^2-4AC}}{2A} = 0, 2x$
Integrating, we find the 2 curves as $y = $constant and $y-x^2 = $constant, so we choose
$$\xi = y,\qquad \eta = y-x^2$$
and apply the relations found by chain rule for differentiating $u$ w.r.t. $\xi$ and $\eta$:
$u_x = u_ξξ_x + u_ηη_x, u_y = u_ξξ_y + u_ηη_y,$
$u_{xx} = u_{ξξ}ξ^2_x + 2u_{ξη}ξ_xη_x + u_{ηη}η^2_x+ u_ξξ_{xx} + u_ηη_{xx}$
$u_{xy} = u_{ξξ}ξ_xξ_y + u_{ξη} (ξ_xη_y + ξ_yη_x) + u_{ηη}η_xη_y + u_ξξ_{xy} + u_ηη_{xy},$
$u_{yy} = u_{ξξ}ξ^2_y + 2u_{ξη}ξ_yη_y + u_{ηη}η^2_y + u_ξξ_{yy} + u_ηη_{yy}$
and substituting in original equation will give $4x^3u_{\xi\eta} = -1$
We want to eliminate $x$ so that equation consists only of $\xi$ and $\eta$, so put $$x^3 = (\xi - \eta)^{3/2}$$ then integrate equation once w.r.t. variable $\xi$ , then with $\eta$ to get the given solution.
Best Answer
The Characteristic equation $$\big(\frac{d y}{d x}\big)^2-2\frac{d y}{d x}+1=0$$ gives only one characteristic curve, $$y-x=constant.$$ So, taking $\xi=y-x$ and $\eta=y$, the given equation reduces to $$u_{\eta \eta}=0.$$