If I have a set of curves in the $xy$-plane with tangent vectors $(1,y)$, how can I conclude that these curves have slope $y$? The context of this question is the PDE $$u_x +yu_y = 0$$
Edit : If the derivation of this result is particularly long or tedious, then please guide me to the relevant theorems/mathematical architecture to which this problem is related.
By definition, for some parametrization $\mathbf{r}(t)=(x(t),y(t))$ of it, the tangent vector of a curve is $$ \left(\frac{\partial x}{\partial t}, \frac{\partial y}{\partial t}\right) $$ Now you are asking about the slope of the function $x \rightarrow f(x)$ whose graph $(x,f(x))$ is identical to $(x(t),y(t))$. That means that $f(x(t))=y(t)$, so by chain rule $$ \frac{\partial y}{\partial t} = \frac{\partial f(x(t))}{\partial t} = \frac{\partial x(t)}{\partial t} \frac{\partial f}{\partial x}(x(t)) $$ For the particular equality that you mention, $$ \left(\frac{\partial x}{\partial t}, \frac{\partial y}{\partial t}\right) = (1,y(t)) $$ you obtain $$ y(t) = 1 \times \frac{\partial f}{\partial x}(x(t)) $$ and thus $f'(x)=y$.