Consider the following PDE, $$ u_x - 2xy^2 u_y = 0 $$ Does there exist a non-trivial solution $u\in \mathcal{C}^1(\mathbb{R}^2,\mathbb{R})$?
It is clear that all solutions for $u\in \mathcal{C}^1( \mathbb{R}^2_+,\mathbb{R})$ are given by $u(x,y) = f\left( x^2 - \tfrac{1}{y}\right)$ where $f\in \mathcal{C}^1(\mathbb{R}_+,\mathbb{R})$. But can we extend such solutions to the entire plane?
The only potential problem is at $y = 0$. For continuity we need $f(t)$ to go to a limit as $t \to \pm \infty$, and then we can define $u(x,0)$ as that limit (let's call it $c$). For partial differentiability wrt $y$ we need existence of $$ \lim_{y \to 0} \dfrac{u(x,y) - u(x,0)}{y} = \lim_{y \to 0} \dfrac{f(x^2 - 1/y) - c}{y} = \lim_{t \to \pm \infty} (x^2 - t) (f(t) - c) $$ For example, you could take $f(t) = \dfrac{1}{t^2+1}$, so that $$ u(x,y) = \dfrac{y^2}{(x^2 y - 1)^2 + y^2} $$ which is $C^\infty$.
