Question: Let $\mathbb{X}$ and $\mathbb{Y}$ be vector fields on $\mathbb{R}^3$ given by $$\mathbb{X}(x,y,z)=(1,0,p(x,y)r(z))$$ $$\mathbb{Y}(x,y,z)=(0,1,q(x,y)r(z))$$ where $p,q$ and $r$ are smooth, and $r$ is nonvanishing. Derive necessary and sufficient conditions on $p,q$ and $r$ so that $$[\mathbb{X},\mathbb{Y}]=0$$
Answer: $[\mathbb{X},\mathbb{Y}]=(\mathbb{X}\cdot\nabla)\mathbb{Y}-(\mathbb{Y}\cdot\nabla)\mathbb{X}$
$$(\mathbb{X}\cdot\nabla)\mathbb{Y}=(dx+p(x,y)r(z)dz)(0,1,q(x,y)r(z))=(0,0,\frac{dq(x,y)}{dx}r(z)+p(x,y)q(x,y)r(z)r'(z))$$$$(\mathbb{Y}\cdot\nabla)\mathbb{X}=(dy+q(x,y)r(z)dz)(1,0,p(x,y)r(z))=(0,0,\frac{dp(x,y)}{dy}r(z)+p(x,y)q(x,y)r(z)r'(z))$$
Therefore for $[\mathbb{X},\mathbb{Y}]=0$ we must have $$\frac{dq(x,y)}{dx}r(z)+p(x,y)q(x,y)r(z)r'(z)=\frac{dp(x,y)}{dy}r(z)+p(x,y)q(x,y)r(z)r'(z)$$$$\Leftrightarrow\frac{dq(x,y)}{dx}r(z)=\frac{dp(x,y)}{dy}r(z)\Leftrightarrow\frac{dq(x,y)}{dx}=\frac{dp(x,y)}{dy}$$
Point of contention: I have found conditions for this to be the case (I think), how do I know that they are necessary and sufficient conditions?
You showed that $[\mathbb{X},\mathbb{Y}]=0 \implies \frac{dq(x,y)}{dx}=\frac{dp(x,y)}{dy} , r(z) \neq 0$. All that you need to show now is that the converse holds: $[\mathbb{X},\mathbb{Y}]=0 \Leftarrow \frac{dq(x,y)}{dx}=\frac{dp(x,y)}{dy} , r(z) \neq 0$.