Q:

Are levelsets of a scalar field without critical points a foliation?

Let $f$ be a differentiable scalar field on an $n$-dimensional Riemannian manifold $X$ without critical points, i.e. $\nabla f \neq 0$ everywhere on $X$. (Assuming $X$ has properties as required to admit such a field.)
Then, by the Frobenius Theorem, the integral curves of $\nabla f$ form a regular foliation on $X$.
My question:
Do the levelsets, i.e. the fibers of $f$ also form a regular foliation on $X$, i.e. with codimension $1$ that is transversal to the foliation given by the integral curves of $\nabla f$?
Intuitively I would think this is trivially true. However I would like to know a proper statement, theorem or simple proof that justifies this assumption. Also required properties of $X$ would be good to know, i.e. besides admitting $f$ as described.
Background:
I am asking this while reasoning about my related question Homotopy of integral curves of a gradient field preserves levelsets (?) I think this is a core ingredient to answer at least some part of that question. Specifically, I think that the regions $U_j$ mentioned in that question are manifolds on which $f$ has no critical points. If the levelsets of $f$ would form a foliation of each $U_j$ this would justify assumption (4) from that question I suppose.

A:

Yes, they do and you can formulate the statement even more general:
Let $M$ be a smooth manifold and $f:M\to \mathbb{R}$ a smooth function such that $d f$ is nowhere vanishing. Then $\ker df$ is a regular foliation. Moreover any vector field $V$ with $df(V)\ne 0$ on $M$ gives rise to a bifoliation on $M$, i.e. $V$ defines a foliation with $TM=\ker df \oplus V$.
To see this take $X,Y\in\ker df$ two vector fields we compute:
$0=d^2 f(X,Y)= X(df(Y))-Y(df(X))-df([X,Y])=df([X,Y])$ 
and hence $[X,Y]\in \ker df$. Since $\ker df$ is moreover of constant rank the Frobenius theorem tells us that it defines a regular foliation.   

