The PDE is $xu_x-yu_y+yu=y$ .
The method of characteristics gives $\dfrac{dx}{x}=-\dfrac{dy}{y}=\dfrac{du}{y-yu}$
Then $x=-c_1y$ and thus $c_1=-\dfrac{x}{y}$ .
Then, I did $du=\dfrac{(y-yu)dy}{y}$ to try to solve for the second constant so that I can have that in terms of $f\left(-\dfrac{x}{y}\right)$ but I can't seem to do that.
$xu_x-yu_y+yu=y$
$yu_y-xu_x=y(u-1)$
$u_y-\dfrac{x}{y}u_x=u-1$
Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:
$\dfrac{dy}{dt}=1$ , letting $y(1)=1$ , we have $y=t$
$\dfrac{dx}{dt}=-\dfrac{x}{y}=-\dfrac{x}{t}$ , letting $x(1)=x_0$ , we have $x=\dfrac{x_0}{t}=\dfrac{x_0}{y}$
$\dfrac{du}{dt}=u-1$ , we have $u(x,y)=f(x_0)e^t+1=f(xy)e^y+1$
$u \equiv w + 1\quad\Longrightarrow yw_{y} - xw_{x} + yw = 0$ $$ w = \phi\left(x\right)\varphi\left(y\right) \quad\Longrightarrow\quad y\varphi'\left(y\right) + y = x\phi'\left(x\right) = \mu = \mbox{constant} $$ $$ y\varphi'\left(y\right) + y = \mu \quad\Longrightarrow\quad \varphi'\left(y\right) = {\mu \over y} - 1 \quad\Longrightarrow\quad \varphi\left(y\right) = \mu\ln\left(y\right) - y + A\,,\ A: \mbox{constant} $$
$$ \phi'\left(x\right) = {\mu \over x} \quad\Longrightarrow\quad \phi\left(x\right) = \mu\ln\left(x\right) + B\,,\quad B: \mbox{constant} $$
$$ \begin{array}{|c|}\hline\\ \color{#ff0000}{\large\quad% u \color{#000000}{\ =\ } 1 + \left[\vphantom{\LARGE A}\mu\ln\left(x\right) + B\right] \left[\vphantom{\LARGE A}\mu\ln\left(y\right) - y + A\right] \quad} \\ \\ \hline \end{array} $$
