Universal differential identities, part 2 This is a follow-up of this question.
Let $$F(f,f_{x_1},f_{x_2},\dots,f_{x_n},f_{x_1x_2},f_{x_1x_3}...)=0$$ be an identity relating some finite number of derivatives of a map $f:\mathbb{R}^n \to \mathbb{R}$. Suppose it is satisfied by all smooth maps.
Must $F$ be the trivial identity, i.e the zero map? 
(Note we already take into account the equality of mixed derivatives, i.e for each pair $x_i,x_j$ we have as an argument only one of 
 the expressions $f_{x_ix_j} \,,\,f_{x_jx_i}$).

The previous question was about the special case where $F$ was a polynomial identity.
The point of the current generalization is to rule out the existence of non-trivial universal identities such as $\sin(f_x)=f_y$. (This particular identity is represented by $F(x_1,x_2,x_3)=\sin x_2-x_3$).
 A: Edit: In general this is false. I will present a special case here and the edit below gives a counterexample. 
I'm not sure what I should assume on $F$ because the word identity is vague, but I will give this a shot. 
Suppose for contradiction that $F(y) = q\neq 0$ at some point vector $y$.
We need to construct a function $f\colon\mathbb{R}^{n}\to\mathbb{R}$ so the the required vector of derivatives at some point is $y$.  To restrict to a nice special case, let us suppose that our identity does not involve any mixed partials. 
Write $h_{j}(x) = \sum_{i=0}^{n} c_{ij}x^{n}$. The first derivative at $x=0$ is $c_{1j}$. The second derivative is $c_{2j}2!$ and the nth derivative is $c_{nj}n!$. Thus, if you want the mth derivative with respect to kth variable to be equal the number $a$, set $c_{mk}=a/m!$ and write $f(x) = \sum_{i=1}^{n} h_{i}(x_{i})$. We pick the $c_{mk}$ so that we get $y$ and then  $$0=F(f,f_{x_1},f_{x_2},\dots,f_{x_n},f_{x_1x_2},f_{x_1x_3}...)(0) = F(f(0),f_{x_1}(0),f_{x_2}(0),\dots,f_{x_n}(0),f_{x_1x_2}(0),f_{x_1x_3}(0)...)=F(y)=q\neq0$$
which is a contradiction.
Now, I will not do the case with mixed partials, but going from the form of a multi-variable Taylor expansion/power series, I imagine that  you can cook something up.
I hope this is clear/helps, but comment if not.
edit:
I just realized something. This is not true if we allow mixed partials. Take for example $F(f_{xy},f_{yx}) = f_{xy}-f_{yx}$. This is $0$ on all smooth functions because all smooth functions have equal mixed partials, but if the mixed partials are not equal then it could be nonzero!
