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I want to calculate $fracpartialpartial vecx fleft(Avecx + vecbright)$ where $vecx, vecb in mathbbR^n$ and $f: mathbbR^n rightarrow mathbbR^n$ which is applied element-wise. Is it correct that since both $vecx$ and $fleft(Avecx + vecbright)$ are vectors the partial derivative must be a Jacobian matrix? I tried applying the chain rule and various identities I have found on Wikipedia, but I am very unsure about the result:
$$fracpartialpartial vecx fleft(Avecx + vecbright) = fracpartialpartial vecx left(Avecx + vecbright)operatornamediagleft(f'left(Avecx + vecbright)right) = Aoperatornamediagleft(f'left(Avecx + vecbright)right)$$

It will be in Jacobian form:

$fracdelta (f_i)delta(x_j)(Ax+b) * A = Jf(Ax+b)*A$

where $f(x_1,...,x_n) = (f_1(x_1,...,x_n),...f_n(x_1,...,x_n))$

I'll assume that you have a specific function in mind, for which you know the derivative,
$$eqalign
fracdf(z)dz &= g(z) cr
df &= g,dz cr
$$
When this function is applied element-wise to a vector argument,
the Hadamard ($circ$) Product must be used in the differential
$$eqalign
df &= gcirc dz cr
$$
For this problem, define
$$eqalign
z &= Ax + b cr
dz &= A,dx cr
$$
then the differential of your function is
$$eqalign
df &= gcirc dz cr
&= G,dz cr
&= GA,dx cr
fracpartial fpartial x &= GA cr
$$
where the matrix
$,G = rm Diag(g)$

Yes, it is a Jacobian matrix. When I get confused with vectors/tensors ("was there a transpose?"), I always start writing down a couple of components and then try to infer what the general expression is.

If you tried but still can't figure it out, read further. Let's compute the derivative of the $i$-th component of $f$ with respect of the $j$-th componente of $x$. To do this, don't forget that every component of $Ax+b$ depends on $x_j$, so we need to use the chain rule $n$ times.

$$
partial_x_j(f_i(Ax+b)) =\
(partial_x_1f_i)(Ax+b)cdot partial_x_j(Ax+b)_1+cdots+(partial_x_nf_i)(Ax+b)cdot partial_x_j(Ax+b)_n\
=(partial_x_1f_i)(Ax+b)cdot partial_x_ja_1j+cdots+(partial_x_nf_i)(Ax+b)cdot partial_x_ja_nj\
=(nabla f_i)(Ax+b)cdot a_j
$$
where $a_j$ is the $j$-th column of $A$. So how does this read when we put all the components together? The entry $J_ij$ of the Jacobian is what we just computed. Well, $J_ij$ is the product of the $i$-th row of $J_f$ with the $j$-th column of $A$. Therefore, denoting products with $*$ to not confuse with function evaluation,
$$
J_f(Ax+b) = J_f(Ax+b)*A.
$$

What is your definition of $fracpartialpartial vec x$?

If we use the definition
$$
fracpartial g partial vec x(vec a) = lim_h to 0 fracg(a+h vec x)-g(a)h,
$$
(for a function $g:mathbb R^n to mathbb R$), then the usual rules of limits should give the answer $A$.