Checking a solution of a PDE I have the following PDE:
\begin{equation}
  -yu_x + xu_y = 0 \quad\text{where } u(0, y) = f(y)
\end{equation}
I derived a solution as follows:
\begin{align}
  -yu_x + xu_y =& 0 \\
  \iff& \nabla u(x,y)\dot \langle -y, x\rangle = 0 \\
  \implies& \frac{dy}{dx} = \frac{-x}{y} \\
  \iff& ydy = -xdx \\
  \iff& \int ydy = \int -xdx \\
  \iff& y^2 = -x^2 + c_2 \\
  \iff& y^2 + x^2 = c_2
\end{align}
Since $u(x,y)$ is constant along the ODE $\frac{dy}{dx}$, we have:
\begin{align}
  u(x,y) =& c_1 \\
         =& f(c_2) \quad\text{for some function $f$} \\
         =& f(y^2+x^2)
\end{align}
I want to check that this satisfies the PDE. Specifically, any function $f$ should satisfy the pde. My calculus is a bit rusty, and I am not exactly sure how to do this. Here is my reasoning:  
Since $u(x,y) = -yu_x + xu_y = 0$ we have to substitute in for $f$ which yields
\begin{equation}
  u(x, y) = -yf_x2x + xf_y2y = 2xyf_y - 2xyf_x
  \end{equation}
We need the above to equal $0$. The $2xy$ and $-2xy$ give me evidence that it should cancel, and that my calculus is off... What is not clear to me is that, we don't know what $f$ is, hence I can't find out what $f_x$ or $f_y$ are. Though, the problem does look symmetrical, and I could see a potential solution involving polar coordinates, but I'm not quite sure how to solve it in this way either.


*

*How can I verify that the above is indeed equal to $0$ and hence satisfies the PDE?
Thanks for all the help!

Since we conjecture $u(x,y)=f(x^2+y^2)$, let us (carefully) apply calculus and verify $-yu_x+xu_y=0$.
Note (by chain rule),
$$ u_x = f'(x^2+y^2)*2x, \qquad u_y = f'(x^2+y^2) * 2y. $$
Thus, $-yu_x+xu_y=-2xyf'+2xyf'=0$.
So, you had it, except a small detail with the chain rule. Another way to have caught the mistake -- $f$ is a function of a single variable, so what does $f_x$ or $f_y$ mean?
