2-Distribution given by the Kernel of a linear functional. Consider the linear functional $w \in (\mathbb{R}^{3})^{*}$, given by $w(x,y,z)=dy-zdx$. Define the 2-Distribution : 
$$ D_{2}:\mathbb{R}^{3}\to T \mathbb{R}^{3}$$ given by 
$$(x,y,z) \to D_{2}(x,y,z)=ker(w(x,y,z))\subset T_{(x,y,z)}\mathbb{R}^{3}_{(x,y,z)}$$
Decide whether or not $D_{2}$ is integrable. 
I decided to find a basis for the $ker(w(x,y,z))$ (due to this subspace has dimension $2$), use the Lie bracket and the Frobenius theorem to say if it is or not integrable, so I found that this distribution may be generated by the two following fields in $T \mathbb{R}^{3}$:
$$X_{1}=\frac{\partial}{\partial x_{1}}x+\frac{\partial}{\partial x_{2}}xz+\frac{\partial}{\partial x_{3}}z,$$
$$X_{2}=\frac{\partial}{\partial x_{1}}\frac{y}{z}+\frac{\partial}{\partial x_{2}}y+\frac{\partial}{\partial x_{3}}z.$$
I'm not really sure if it's correct  and I haven't been able to prove that this two fields are linearly independent, 
thanks in advanced.
 A: $dw=-dz\wedge dx$, we deduce that $w\wedge dw=-(dy-dz)\wedge dz\wedge dx=-dy\wedge dz\wedge dx\neq 0$ and the distribution is not integrable from the formulation of Frobenius for $1$-forms.
Frobenius theorem for differential forms
A: The other answer covers it, but we can do it from scratch as follows: $(dy-zdx)\left(a\frac{\partial }{\partial x}+b\frac{\partial }{\partial y}+c\frac{\partial }{\partial z}\right)=0\Leftrightarrow b-za=0$ so $X=\frac{\partial }{\partial x}+z\frac{\partial }{\partial y}$ and $Y=\frac{\partial }{\partial z}$ are basis elements for the nullspace of $\omega.$ Now, with $x=x^1,y=x^2,z=x^3$ we have
$[X,Y]=\left(X^i\frac{\partial Y^j}{\partial x^i}-Y^i\frac{\partial X^j}{\partial x^i}\right)\frac{\partial}{\partial x^j}=\left(\frac{\partial Y^j}{\partial x^1}+z\frac{\partial Y^j}{\partial x^2}-\frac{\partial X^j}{\partial x^3}\right)\frac{\partial}{\partial x^j}=(0+0+0)\frac{\partial}{\partial x^1}+(0+0-1)\frac{\partial}{\partial x^2}+(0+0+0)\frac{\partial}{\partial x^3}=-\frac{\partial}{\partial y}.$
Now, at any point $p=(x,y,z)\in \mathbb R^3$ such that $z=0,\ \frac{\partial}{\partial y}\notin \text{span}\ \{X,Y\}$  so $D$ is not integrable.
