Given: $T = \langle x\partial_z - z\partial_y, z\partial_x - x\partial_z\rangle$.
First we should check whether your given fields are indeed linearly independent. This is true if and only if $z \neq 0$, because this is precisely when $$\begin{bmatrix} 0 & z \\ -z & 0 \\ x & -x\end{bmatrix}$$ has full rank. Meaning that $T$ is a distribution on $V$. Then you should check whether $T$ is indeed involutive. We have $$\begin{align}[x\partial_z-z\partial_y, z\partial_x - x\partial_z] &= [x\partial_z, z\partial_x] - [x\partial_z,x\partial_z] - [z\partial_y, z\partial_x]+[z\partial_y, x\partial_z] \\ &= (x\partial_x - z\partial_z)- 0 - 0- x\partial_y \\ &= x\partial_x - x\partial_y - z\partial_z.\end{align}$$Writing $$(x,-x,-z) = (0,-zf,xf) + (zg,0,-xg)$$leads to the system $$\begin{cases}x = zg \\ -x = -zf \\ -z = xf-xg, \end{cases}$$which has no solution ($f = g= x/z$ is not compatible with the last equation).
So unless I have made any computation mistake, your distribution $T$ is not involutive, hence not integrable, and you will not be able to find flat charts.
For the edited version, we were given $T= \langle y\partial_z - z\partial_y, z\partial_x - x\partial_z\rangle$.
We start following the same recipe. The matrix $$\begin{bmatrix}0 & z\\ -z & 0 \\ y & -x \end{bmatrix}$$again has full rank if $z \neq 0$, which happens since all the points in $V$ have positive coordinates, so we indeed have a distribution on $V$. Next, we compute the Lie bracket to check whether $T$ is involutive or not: $$\begin{align} [y\partial_z - z\partial_y, z\partial_x - x\partial_z] &=[y\partial_z, z\partial_x] - [y\partial_z, x\partial_z] - [z\partial_y, z\partial_x]+[z\partial_y, x\partial_z] \\ &= y\partial_x-0-0-x\partial_y \\ &=y\partial_x - x\partial_y.\end{align}$$Writing $$(y,-x,0) = (0,-zf,yf)+(zg,0,-xg)$$leads to the system $$\begin{cases} y = zg \\ -x=-zf \\ 0 = yf-xg,\end{cases}$$which has the solution $f=x/z$ and $g=y/z$. So $T$ is involutive and hence integrable, so we will be able to find flat charts for $T$. The idea here is that if $X$ and $Y$ are commuting vector fields spanning $T$, the combined flow $$(t,s)\mapsto \Phi_{t,X}\circ \Phi_{s,Y}(x,y,z)$$gives a parametrization for the integral surface of $T$ passing through the point $(x,y,z)$. This understood, the idea here is to first find such fields. Since the first $2\times 2$ submatrix of the coefficient matrix above has full rank, one can just do $$\begin{bmatrix} X \\ Y \end{bmatrix}
= \left(\begin{bmatrix} 0 & -z \\ z & 0\end{bmatrix}^{-1}\right)^\top \begin{bmatrix} y\partial_z-z\partial_y \\ z\partial_x -x\partial_z\end{bmatrix}.$$Once you have the expressions for $X$ and $Y$ you can solve two systems of ODEs to find the flows, and from this get the flat chart. Can you go on?
More comments: I kept thinking about this in the last couple of days, and indeed the algorithm above is not something natural to think of. So here's an easier way to think of it: using the formulas from classic vector calculus, you can prove that if ${\bf N}$ is a non-vanishing vector field along some open set $U\subseteq \Bbb R^3$, then ${\bf N}^\perp \hookrightarrow TU$ is involutive if and only if $\langle {\bf N}, {\rm curl}({\bf N})\rangle =0$. Taking the cross product, we can describe $T$ as ${\bf N}^\perp$ for $\mathbf{N} = xz\partial_y+yz\partial_y+z^2\partial_z$. Then ${\rm curl}({\bf N}) = -y\partial_x+x\partial_y$ is orthogonal to ${\bf N}$ and so $T$ is involutive. Since $z>0$ on $V$, we may describe $T$ by the relation $${\rm d}z = -\frac{x}{z}\,{\rm d}x - \frac{y}{z}\,{\rm d}y.$$Now, we have that $$\left({\rm d}x, {\rm d}y, {\rm d}z +\frac{x}{z}\,{\rm d}x+\frac{y}{z}\,{\rm d}z \right)$$is a coframe for $T^*V$, and dualizing this back to a frame on $TV$ we get $$\left(\partial_x - \frac{x}{z}\partial_z, \partial_y - \frac{y}{z}\partial_z, \partial_z\right).$$By linear algebra, the first two fields give a trivialization for $T$, and they automatically commute because we have already checked that $T$ is involutive (in fact, they commute if and only if $T$ is involutive -- see here). Thinking of it like this we avoid paranoia with submatrices and keeping track of whether coordinates should be permuted or not.
Your solution is almost correct, at least your computations are. You should have conclude that $\partial_xf\equiv 0$, so that $f=f(y,z)$ is a function of $y$ and $z$ only.
Perhaps another way to conclude would have been to use the differential form version of Frobenius theorem, stating that $D=\ker\alpha$ is integrable if, and only if, $\alpha\wedge\operatorname{d}\alpha=0$. In your case, $\alpha$ can be taken to be $\operatorname{d}z-f\operatorname{d}y$, so that $\alpha\wedge\operatorname{d}\alpha=-\operatorname{d}f\wedge\operatorname{d}y\wedge\operatorname{d}z=-\partial_xf\operatorname{d}x\wedge\operatorname{d}y\wedge\operatorname{d}z$, and the same conclusion follows.
Best Answer
In order to have $Y= \frac{\partial}{\partial u}$ for some coordinate system $\lbrace u, v, w\rbrace$, you need to find (independent) functions such that $Y(u) =1, Y(v) = 0= Y(w) $.
You can take $u= x, v= y$, and $w$ will be some other function with $Y(w) =0$. By guessing, you can see that $w= xy-z$ is such function.
In this same system of coordinates, your $\tilde{X}$ is equal to $\frac{\partial}{\partial v}$.