Function never equal to 0 Let $p,q,x,y:(a,b)\to\mathbb{R}$ be $C^1 ((a,b))$ functions. Knowing that for  $(u,v), (\tilde{u},\tilde{v})\in (a,b)\times (0,1)$ we have that:

$$\begin{cases} p(u)+vx(u)=p(\tilde{u})+\tilde{v}x(\tilde{u}) \\ q(u)+vy(u)=q(\tilde{u})+\tilde{v}y(\tilde{u}) \end{cases}\ \Longleftrightarrow\ (u,v)=(\tilde{u},\tilde{v}),$$

is it true that the function:

$f:(a,b)\times (0,1)\to\mathbb{R}$ given by:

$f(u,v)=p'(u)y(u)-q'(u)x(u)+v(x'(u)y(u)-x(u)y'(u))$, is NEVER equal to $0$?
Leg $r=(p,q)$ and $z=(x,y)$. Define $h(u,v):=r(u)+vz(u)$.  From the given equivalence we know that $h$ is injective.  Now verify that $f$ is the determinant of the Jacobian of $h$ which is doomed not to be zero because $h$ is injective.
Leg $r=(p,q)$ and $z=(x,y)$. Define $h(u,v):=r(u)+vz(u)$.  From the given equivalence we know that $h$ is injective.  Now verify that $f$ is the determinant of the Jacobian of $h$ which is doomed not to be zero because $h$ is injective.