How to find the integral submanifold? Suppose $U \subseteq R^3$ is the subset that all three coordinates are positive. Let $D$ be the distribution on $U$ spanned by two vector fields:  
$X = y\frac{\partial }{\partial z}-z\frac{\partial }{\partial y}$ , $Y=z\frac{\partial }{\partial x}-x\frac{\partial }{\partial z}$.  
Then how to find the integral submanifold?  
I had shown that $[X, Y]_p = \frac{y(p)}{z(p)}Y_p+\frac{x(p)}{z(p)}X_p$, hence $D$ is involutive. I also know how to calculate the integral curve of $X$ and $Y$. But I dont know how to proceed to calculate the integral submanifold specifically.
 A: In this case, the result follows by close inspection of the two vector fields: one generates rotation in the $y,z$-plane and the other generates rotation in the $x,z$-plane.
This is a good point to pause and think geometrically about what the integral submanifolds of these two vector fields might be.
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Using this observation, you should be able to ``see'' geometrically that the integral submanifolds for your distribution are given as level sets of the function
$$ f(x,y,z) = x^2 + y^2 + z^2.$$
Let's prove this in detail.  The level sets of the function $f$ are all 2-dimensional (except for the fiber $f^{-1}(0) = {(0,0,0)}$, which we ignore for the moment). They are concentric spheres.  
The foliation of $\mathbb{R}^3$ by level sets of $f$ has an involutive tangent distribution (of dimension 2, since they are regular fibers of a smooth map $\mathbb{R}^3 \to \mathbb{R}^1$). At each point, this distribution is defined as the kernel of the derivative of $f$,
$$ \ker(Df_p).$$
You can check that 
$$ X\vert_p, Y\vert_p \in \ker(Df_p)$$
(or equivalently, you can check that the Lie derivatives $L_X f = L_Y f = 0$).
Thus, your distribution is contained in the distribution $\ker(Df_p)$. But they both have dimension 2, so they coincide. 
Thus the integral submanifolds for your distribution are the same as the level sets of $f$: they are the concentric spheres centered at the origin.
If you want a nice parameterization of this foliation, you are in luck, it has been studied a lot. Spherical or cylindrical coordinates are common favourites.
