First, we have \[ au_{xx} + 2b u_{xy} + c u_{yy} = 0, \]

\noindent where \begin{gather*} a = y \\ b = \frac{x + y}{2} \\ c = x. \end{gather*}

The goal is for us to use a general coordinate transformation $(x,y) = (\xi, \eta)$, so that the differential equation becomes now \[ A u_{\xi\xi} + 2B u_{\xi \eta} + C u_{\eta\eta} + \Phi = 0, \]

and the coefficients are \begin{align*} \begin{pmatrix} A & B \\ C & D \end{pmatrix} \begin{pmatrix} \xi_x & \xi_y \\ \eta_x & \eta_y \end{pmatrix} \begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} \xi_x & \xi_y \\ \eta_x & \eta_y \end{pmatrix}. \end{align*}

The key behind all this classification guff is that the coefficients are related through \[ a^2 - bc = [A^2 - BC] \frac{\partial(\xi, \eta)}{\partial(x,y)}, \]

so that the discriminant never changes sign regardless of how we perform the transformation. By lectures, you know that the PDE is hyperbolic if \[ a^2 - bc = (x - y)^2 > 0, \]

and thus the problem is hyperbolic everywhere except along $y = x$.

Now we must reduce it to canonical form, \[ Bu_{\xi \eta} + \Phi = 0. \]

(If you are confused as to why that is hyperbolic-looking, try setting $X = \xi + \eta$ and $Y = \xi - \eta$ to derive a wave equation). We thus need to set $A = C = 0$. This amounts to solving the pair of PDEs \begin{align*} a \xi_x^2 + 2b \xi_x \xi_y + c \xi_y^2 &= 0 a \eta_x^2 + 2b \eta_x \eta_y + c \eta_y^2 &= 0, \end{align*}

which, by factorization, amounts to solving \begin{align*} \xi_x + \lambda_1 \xi_y &= 0 \\ \eta_x + \lambda_2 \eta_y &= 0 \end{align*}

where $\lambda_1$ and $\lambda_2$ are solutions to $a\lambda^2 - 2b\lambda + c = 0$. The quadratic formula gives us \[ \lambda = \frac{1}{2y} [(x + y) \pm (x-y)] = 1, \ \frac{x}{y}. \]

The characteristics of $\xi_x + \xi_y = 0$ are given by $y = x + C$, and since $\xi$ is constant along characteristics, then the solution is simply $\xi = h(y - x)$ where for the simplest coordinate transform, you can take $\xi = y - x$. Similarly, the characteristics for $\eta_x + (x/y) \eta_y = 0$ are $y^2 - x^2 = C$, and so we can choose $\eta = y^2 - x^2$.

Now the painful part. We must obtain $\Phi$. For this, we must use the chain rule to obtain $u_{xx}$, $u_{xy}$, and $u_{yy}$. There is no easy way to do this except to barrel through. In the end, you should have \[ \xi u_{\xi \eta} + u_\eta = 0. \]

Integrating with respect to $\xi$ gives \[ \xi u_\eta = f'(\eta), \]

for some unknown function, $f(\eta)$. Integrating now in $\eta$ and rerranging gives \[ u = \frac{f(\eta)}{\xi} + g(\xi), \]

for some function $g(\xi)$. Finally, using $\xi = y - x$ and $\eta = y^2 - x^2$, we have \[ u(x,y) = \frac{f(y^2 - x^2)}{y-x} + g(y - x). \]