Characteristic equation
$A(\dfrac{dy}{dx})^2-B(\dfrac{dy}{dx})+C=0\Rightarrow y(\dfrac{dy}{dx})^2-(x+y)(\dfrac{dy}{dx})+x=0$
then
$\dfrac{dy}{dx}=\dfrac{(x+y)\pm\sqrt{(x-y)^2}}{2y}=\dfrac{x}{y},\ 1
$
Characteristic curves: $s(x,y)$ and $t(x,y)$
$\left\{\begin{array}{c}
\displaystyle{\frac{dy}{dx}=\frac{x}{y}} \\
\displaystyle{\frac{dy}{dx}=1}
\end{array}\right.\ \Rightarrow\
\left\{\begin{array}{l}
s(x,y)=y^2-x^2=c_1\ \\
t(x,y)=y-x=c_2
\end{array}\right.
$
Reduce to Standard Form by using characteristic curves with chain rules
(i) First order derivatives
$
\begin{align*}
u_{x}=\frac{\partial u}{\partial x}&=\frac{\partial u}{\partial s}\frac{\partial s}{\partial x}+\frac{\partial u}{\partial t}\frac{\partial t}{\partial x}=s_x u_s+t_x u_t
\\
u_{y}=\frac{\partial u}{\partial y}&=\frac{\partial u}{\partial s}\frac{\partial s}{\partial y}+\frac{\partial u}{\partial t}\frac{\partial t}{\partial y}=s_y u_s+t_y u_t
\end{align*}$
(ii) Second order derivatives
$
\begin{align*}
u_{xx}&=\frac{\partial }{\partial x}(s_x u_s+t_x u_t)=\Big(s_{xx}u_s+t_{xx}u_t\Big)+\Big(s_x\frac{\partial u_s}{\partial x}+t_x\frac{\partial u_t}{\partial x}\Big)
\\
&=\Big(s_{xx}u_s+t_{xx}u_t\Big)+s_x\Big(s_xu_{ss}+t_xu_{st}\Big)+t_x\Big(s_xu_{ts}+t_xu_{tt}\Big)\\
&=s_{xx}u_s+t_{xx}u_t+s_x^2u_{ss}+2s_xt_xu_{st}+t_x^2u_{tt}
\\
u_{yy}&=s_{yy}u_s+t_{yy}u_t+s_y^2u_{ss}+2s_yt_yu_{st}+t_y^2u_{tt}
\\
u_{xy}&=\frac{\partial }{\partial y}(s_x u_s+t_x u_t)=\Big(s_{xy}u_s+t_{xy}u_t\Big)+\Big(s_x\frac{\partial u_s}{\partial y}+t_x\frac{\partial u_t}{\partial y}\Big)
\\
&=\Big(s_{xy}u_s+t_{xy}u_t\Big)+s_x\Big(s_yu_{ss}+t_yu_{st}\Big)+t_x\Big(s_yu_{ts}+t_yu_{tt}\Big)\\
&=s_{xy}u_s+t_{xy}u_t+s_xs_yu_{ss}+(s_xt_y+s_yt_x)u_{st}+t_xt_yu_{tt}
\end{align*}
$
(iii) Substitute for $u_{xx}, u_{yy}$ and $u_{xy}$ in the PDE:$y u_{xx}+(x+y)u_{xy}+x u_{yy}=0$.
We can get
$
\begin{align*}
y&\Big[(-2)u_{s}+(-2x)^2u_{ss}+2(-2x)(-1)u_{st}+(-1)^2u_{tt}\Big]\\
&+(x+y)\Big[(-2x)(-2y)u_{ss}-2(x+y)u_{st}+(-1)(1)u_{tt}\Big]\\
&+x\Big[(2)u_{s}+(2y)^2u_{ss}+2(2y)(1)u_{st}+(1)^2u_{tt}\Big]=0
\end{align*}$
rearrange the terms
$-2(y-x)u_{s}-2(y-x)u_{st}=0\Rightarrow u_{s}+tu_{st}=0
$
For $y\neq x\Rightarrow$
$\dfrac{1}{t}\dfrac{\partial u}{\partial s}+\dfrac{\partial}{\partial t}(\dfrac{\partial u}{\partial s})=0.
$
Let $z=\dfrac{\partial u}{\partial s}$, then
$\dfrac{1}{t}+\dfrac{1}{z}\dfrac{\partial z}{\partial t}=0
$
integrate
$\ln{t}+\ln{z}=\ln{f(s)}\Rightarrow z=\dfrac{\partial u}{\partial s}=\dfrac{1}{t}f(s)
$
integrate again
$u(x,y)=\dfrac{1}{t}F(s)+g(t)=\dfrac{1}{y-x}F(y^2-x^2)+g(y-x)
$
Another method
By observing, we can express the PDE in
$(y\dfrac{\partial}{\partial x}+x\dfrac{\partial}{\partial y})(\dfrac{\partial}{\partial x}+\dfrac{\partial}{\partial y})u=0
$
Let $z=(\dfrac{\partial}{\partial x}+\dfrac{\partial}{\partial y})u$, then $y\dfrac{\partial z}{\partial x}+x\dfrac{\partial z}{\partial y}=0
$
Characteristic equation
$\dfrac{dx}{y}=\dfrac{dy}{x}=\dfrac{dz}{0}\Rightarrow\left\{\begin{array}{c}
\displaystyle{\frac{dy}{dx}=\frac{x}{y}} \\
\displaystyle{\frac{dz}{dx}=0}
\end{array}\right.\Rightarrow\left\{\begin{array}{l}
\displaystyle{y^2-x^2=c_1} \\
\displaystyle{z=c_2}
\end{array}\right.$
$\because c_2=f(c_1), z=f(y^2-x^2)$
$\Rightarrow f(y^2-x^2)=\dfrac{\partial u}{\partial x}+\dfrac{\partial u}{\partial y}$
Characteristic equation again
$\dfrac{dx}{1}=\dfrac{dy}{1}=\dfrac{du}{f(y^2-x^2)}\Rightarrow\left\{\begin{array}{l}
\displaystyle{\frac{dy}{dx}=1} \\
\displaystyle{\frac{du}{d(y+x)}=f(y^2-x^2)}
\end{array}\right.\Rightarrow\left\{\begin{array}{l}
\displaystyle{y-x=c_3} \\
\displaystyle{u=\frac{1}{y-x}F(y^2-x^2)+c_4}
\end{array}\right.$
$\because c_4=f(c_3)$, we finally obtain the solution
$u(x,y)=\dfrac{1}{y-x}F(y^2-x^2)+g(y-x)
$