a) This is a first order linear inhomogeneous PDE. We begin by examining characteristic lines:

\begin{equation*}

\frac{\,dx}{y}=\frac{\,dy}{-x}=\frac{\,du}{xy}

\end{equation*}

First equation implies $x^2+y^2=C$, for some constant $C$. Therefore $u(x,y)=\phi(x^2+y^2)$, for some arbitrary $\phi$ is solution to homogeneous equation. From second equation we get $ \,dx=\frac{\,du}{x}$ which implies $u=\frac{x^2}{2}$ is a particular solution to the inhomogeneous equation.

The general solution would be:

$$u(x,y)=\phi(x^2+y^2)+\frac{x^2}{2}$$

where $\phi$ is arbitrary.

b) General solution to the homogeneous equation is identical to (a). Remaining task is to solve for $u(x,y)$ in $\frac{\,dx}{y}=\frac{\,du}{x^2+y^2}$ and find a particular solution. Substituting $y^2$ with $C-x^2$ we get the following ODE:

$$

C\frac{\,dx}{\sqrt{C-x^2}}=\,du

$$

Integerating both sides we get

\begin{equation*}

u=\int C\frac{\,dx}{\sqrt{C-x^2}} \end{equation*}

$$

=C\int \frac{\,dx}{\sqrt{C}\sqrt{1-(\frac{x}{\sqrt{C}})^2}} \\

=C\arcsin{\frac{x}{\sqrt{C}}} \\

= (x^2+y^2)\arcsin{\frac{x}{\sqrt{x^2+y^2}}}

$$

General solution is

$$u(x,y)=\phi(x^2+y^2)+(x^2+y^2)\arcsin{\frac{x}{\sqrt{x^2+y^2}}}$$

where $\phi: \mathbb{R} \rightarrow \mathbb{R}$ is an arbitrary function.

c) In the latter case, if $(x,y)=(0,0)$ is in domain of $u$, general solution does not exist as $\arcsin{\frac{x}{\sqrt{x^2+y^2}}}$ is not well-defined.