I am confused as to how they got from the two equations being equal to 0 to the derivative being 0. I could be really tired right now but this isn't really making sense to me. I was thinking of doing cases where u isn't 0 and v isn't 0 separately and they assume both aren't 0 and then just solve it out. I feel like I could brute force this but is there a shorter way of doing this without these cases.
If $a^2+b^2\neq0$ the following system : $$\begin{align}ax-by=0\\bx+ay=0\end{align}$$ have the only solution $(0,0)$, because the matrix $$A=\begin{pmatrix}a &-b\\ b &a\end{pmatrix}$$ is invertibe and $\det(A)=a^2+b^2$, So if $X=\begin{pmatrix}x\\ y\end{pmatrix}$ the system is equivalent to $AX=0$ and $A$ invertible implies $X=0$
Note that you can multiply these equations by $u, v$, respectively to $u^2u_x - uvu_y = 0$ and $ uvu_y + v^2 u_x = 0$. Then add them to get that $(u^2+v^2)u_x = 0$, so that $u_x = 0$ (Here we use $c \ne 0$). Getting from here to $u_y = 0$ is then obvious.
