Consider the mapping $\mathbb{R}^3 (x,y,z)$ to $\mathbb{R}^3(u,v,w)$ given in coordinates Consider the mapping $\mathbb{R}^3 (x,y,z)$ to $\mathbb{R}^3(u,v,w)$ given in coordinates
$$
\left\{
\begin{array}{c}
u = yz\sin(x)\\ 
v = y^2 - x\\
w = xz
\end{array}
\right.
$$
Determine the ranks of all points in the domain $\mathbb{R}^3(x,y,z)$
Let $f(x,y,z) = (yz\sin(x), y^2-x, xz)$
so $f(x,y,z) = (u,v,w)$
$$
Df(x,y,z) = \left[
\begin{array}{c}
\cos(x)yz &\sin(x)z & \sin(x)y\\ 
-1 & 2y & 0\\
z & 0 & x
\end{array}
\right]
$$
$= (\cos(x)yz)(2xy) - (\sin(x)z)(-x) + (\sin(x)y)(-2yz)$ 
$= 2x\cos(x)y^2z + (x^2 - 2y^2z)\sin(x)$
That's what I've done so far, I'm stuck on what to do after.
 A: You have computed the Jacobian
$$J(x,y,z)=2xy^2 z\cos x+xz\sin x-2y^2 z\sin x=z\left(
2y^2(x\cos x-\sin x)+x\sin x\right)\ .$$
At the points $(x,y,z)$ where $J(x,y,z)\ne0$ the derivative $df$ has rank $3$; these are  regular points of $f$.
The zero set $C$ of $J$ consists of the plane $z=0$ and of the vertical stalks erected on the zero set $S$  of the function $$g(x,y):=2y^2(x\cos x-\sin x)+x\sin x$$
(this set $S$ is a union of curves in the $(x,y)$-plane).
In the points of $C$ the derivative $df$ has rank $\leq2$. Now rank $=0$ is excluded since the matrix $[df]$ has an entry which is never zero.
Therefore it remains to check whether there rank $=1$ points at all. At such points the  column space of $[df]$ has to be one-dimensional, which implies that the second and third columns have to be multiples of the first column. It follows that the third column is zero, from which we conclude that necessarily $x=0$. Now
$$[df(0,y,z)]=\left[\matrix{yz&0&0\cr -1&2y&0\cr z&0&0\cr}\right]\ .$$
This matrix has rank $2$ when $yz\ne0$ and rank $1$ otherwise. It follows that on the $y$-axis and on the $z$-axis the derivative $df$ has rank $1$, and in all other points of $C$ rank 2.
