Using polar co-ordinates to solve first order linear PDE 
Let $u = u(x,y)$ be a scalar fields on $\mathbb R^2$. 
  Consider the linear first order PDE for $u$,$$(x\partial_y - y\partial_x)u = 0$$
  By introducing $\partial_{\phi}u = 0$planar polar co-ordinates $r$ and $\phi$, show that the PDE is equivalent to $\partial_{\phi}u = 0$, and thus find its general solution.

Ok so far I have tried to find its general solution first:
First,  rewrite the above PDE as
\begin{align}
\partial_x u+\frac{y}{x}\partial_y u=0 \ \ (1)
\end{align}
Using the method of characteristic,  further rewrite $(1)$ as follows
\begin{align}
\frac{d}{dx}u(x, y(x))= \partial_x u+ y'(x)\partial_y u = \partial_x u+\frac{y}{x}\partial_y u=0, 
\end{align}
we have the ode
\begin{align}
y' = \frac{y}{x} \ \ \Rightarrow \ \ y = Cx.
\end{align}
Let us impose the artifical initial condition $y(1) = y_0$ and $u(1, y_0) = f(y_0)$ . Hence it follows
\begin{align}
y = y_0x \ \ \Rightarrow \ \ y_0 = \frac{y}{x}.
\end{align}
which means
\begin{align}
u(x, y) = u\left(x, y_0x\right) = f(y_0) = f\left(\frac{y}{x} \right).
\end{align}
I am having trouble using planar polar co-ordinates $r$ and $\phi$ to show that the PDE is equivalent to $\partial_{\phi} u = 0$. Any help to solve that bit of the problem will be appreciated.
 A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

$\ds{x = r\cos\pars{\theta}\,,\quad y = r\sin\pars{\theta}}$.

\begin{align}
\partiald{}{r} & =
\partiald{x}{r}\,\partiald{}{x} + \partiald{y}{r}\,\partiald{}{y} =
\cos\pars{\theta}\,\partiald{}{x} + \sin\pars{\theta}\,\partiald{}{y} 
\\[5mm]
\partiald{}{\theta} & =
\partiald{x}{\theta}\,\partiald{}{x} + \partiald{y}{\theta}\,\partiald{}{y} =
-r\sin\pars{\theta}\,\partiald{}{x} + r\cos\pars{\theta}\,\partiald{}{y} 
\end{align}

$$
\left\{\begin{array}{rcrcl}
\ds{\cos\pars{\theta}\,\partiald{}{x}} & \ds{+} &
\ds{\sin\pars{\theta}\,\partiald{}{y}} & \ds{=} & \ds{\partiald{}{r}}
\\
\ds{-r\sin\pars{\theta}\,\partiald{}{x}} & \ds{+} &
\ds{r\cos\pars{\theta}\,\partiald{}{y}} & \ds{=} & \ds{\partiald{}{\theta}}
\end{array}\right.
$$

$$
\left\{\begin{array}{rcrcl}
\ds{\partiald{}{x}} & \ds{=} &
\ds{\cos\pars{\theta}\,\partiald{}{r}} & \ds{-} &
\ds{{\sin\pars{\theta} \over r}\,\partiald{}{\theta}}
\\[2mm]
\ds{\partiald{}{y}} & \ds{=} &
\ds{\sin\pars{\theta}\,\partiald{}{r}} & \ds{+} &
\ds{{\cos\pars{\theta} \over r}\,\partiald{}{\theta}}
\end{array}\right.
$$

\begin{align}
&\left\{\begin{array}{rcrcl}
\ds{y\,\partiald{}{x}} & \ds{=} &
\ds{{1 \over 2}\,r\sin\pars{2\theta}\,\partiald{}{r}} & \ds{-} &
\ds{\sin^{2}\pars{\theta}\,\partiald{}{\theta}}
\\[2mm]
\ds{x\,\partiald{}{y}} & \ds{=} &
\ds{{1 \over 2}\,r\sin\pars{2\theta}\,\partiald{}{r}} & \ds{+} &
\ds{\cos^{2}\pars{\theta}\,\partiald{}{\theta}}
\end{array}\right.
\\[5mm] \implies &\
\bbox[10px,#ffe,border:1px dotted navy]{\ds{%
x\,\partiald{}{y} - y\,\partiald{}{x} = \partiald{}{\theta}}}
\quad \implies \quad
\bbox[10px,#ffe,border:1px dotted navy]{\ds{%
\partiald{u}{\theta} = 0}}
\end{align}
