With regards to problem 4 (c)...

Let me begin with an example. Consider the function

$$f(x) = \begin{cases}x^2 \sin\left(\frac{1}{x}\right) & x \ne 0, \\ 0 & x = 0. \end{cases}$$

Although not immediately obvious, it is a simple exercise to show that this function is differentiable at all points on the real line. Indeed, when $x \ne 0$ the "problem term" $\frac{1}{x}$ is differentiable, and so we are free to apply the chain/product rule: $$f'(x) = 2 x \sin\left(\frac{1}{x}\right) - \cos\left(\frac{1}{x}\right)$$

The naive student might conclude that the derivative does not exist at $x = 0$ because this formula no longer makes sense in that region. On the contrary, if we appeal to the definition of a derivative we see that

$$f'(0) = \lim_{h \to 0} \frac{h^2 \sin\left(\frac{1}{h}\right)}{h} = \lim_{h \to 0} h \sin \left(\frac{1}{h}\right) = 0$$

by the squeeze theorem.

A similar fallacy is appearing in your solutions to problem 4 (c). In short, the failure of the method of characteristics, while certainly does indicate that some interesting phenomenon is occurring, does **not** imply that there isn't *some other* method which may be able to solve the problem. In particular, the main problem with 4 (b) is not so much the vanishing of the denominator in the arcsin/arctan term, but rather the fact that it is multivalued.

Here is a quick proof sketch that there is no "global" solution to 4 (b): Consider the circle $x^2 + y^2 = 1$. Rewriting the PDE as $$(x, -y) \cdot \nabla u = x^2 + y^2$$ we see that the characteristics curves are circles. Suppose we begin our journey at the point $(1, 0) \in \mathbb{R}^2$. As we move along the characteristic curve (ie. the circle of radius one) we see from the PDE that the function $u(x,y)$ is increasing at a rate of $x^2 + y^2$ which is exactly $1$ because we are moving along the unit circle! Hence, after a full revolution we will end up with a contradiction that $u(1, 0) = u(1, 0) + 2 \pi$, where the $2 \pi$ comes from integrating the constant $1$ around the unit circle (aka the circumference).

On the other hand, if we're careful about our domain, the method of characteristics can work. For example, you could solve an IVP of 4 (b) on the *upper right quadrant* if I gave you the initial conditions on the positive x-axis, ie. $u(x, 0) = \phi(x)$ for $x > 0$.

To summarize, if your characteristics curves "loop back on themselves" you need to ensure that the inhomogeneous term of the equation is "compatible" with curves. The function $xy$ worked, but $x^2 + y^2$ didn't.