For a scalar argument, the derivatives/differentials of the function are
$$\eqalign{
\sigma'(z_k) &= \frac{d\sigma}{dz_k}\quad\implies\quad d\sigma = \sigma'dz_k \\
\sigma''(z_k) &= \frac{d\sigma'}{dz_k}\quad\implies\quad d\sigma' = \sigma''dz_k \\
}$$
When applied element-wise to vectors, these evaluate to vectors
$$s = \sigma(z),\quad s'=\sigma'(z),\quad s''=\sigma''(z)$$
and the differentials must be written using the element-wise (aka Hadamard) product
$$ds = s'\odot dz,\qquad ds' = s''\odot dz$$
Since the time derivatives/differentials of $z$ are related by
$$dz = \dot z dt,\qquad d\dot z = \ddot z dt$$
the time derivatives of $s$ can be calculated as
$$\eqalign{
ds &= s'\odot\dot z \, dt \\
\dot s &= s'\odot\dot z \\\\
d\dot s
&= ds'\odot\dot z + s'\odot d\dot z \\
&= (s''\odot dz)\odot\dot z + s'\odot(\ddot z dt) \\
&= (s''\odot\dot z\odot\dot z + s'\odot\ddot z)\,dt \\
\ddot s &= s''\odot\dot z\odot\dot z + s'\odot\ddot z \\
}$$
One final trick, is that Hadamard multiplication by a vector can be replaced by first converting the vector into a diagonal matrix and then performing normal matrix multiplication, i.e.
$$\eqalign{
a\odot b &= {\rm Diag}(a)\,b \;=\; Ab \\
{\rm Diag}(a\odot b) &= AB = BA \quad
\big({\rm diagonal\,matrices\,commute}\big) \\
}$$
So these results can be written using diagonal matrices as
$$\eqalign{
\dot S &= S'\dot Z \\
\ddot S &= S''\dot Z^2 + S'\ddot Z \\
}$$
or as a mixture of matrices and vectors
$$\eqalign{
\dot s &= \dot Zs' \\
\ddot s &= \dot Z^2s'' + \ddot Zs' \\
}$$
NB: In the case of the Logistic function the derivatives are given by simple formulas
$$\eqalign{
\sigma' &= (\sigma-\sigma^2)
&\implies S'=(S-S^2) &\implies s'=(I-S)s \\
\sigma'' &= (\sigma-3\sigma^2+2\sigma^3)\;
&\implies S''=(S-3S^2+2S^3)\;
&\implies s''=(I-3S+2S^2)s \\
}$$
The issue is that you are using $x$ as both as a free variable, and as a function, $x(t)$. And likewise $y$.
The notation $\dfrac{\partial f}{\partial t}$ is really being used as an implicit shorthand for $\dfrac{\partial f(x(t),y(t))}{\partial t}$.
$~$ Where $f(x(t),y(t))$ is a convolution of the bivariate function $f$, with monovariate functions $x$ and $y$, each evaluated with the same argument $t$, a free variable.
Likewise you are using the notation $\dfrac{\partial f}{\partial x}$ is used as shorthand for $\left.\dfrac{\partial f(u,v)}{\partial u}\right\vert_{\raise{2ex}{u:=x(t)\\v:=y(t)}}$ .
$~$ That means to evaluate the partial differential of $f$ with respect to its first argument and then form a composition by substituting those arguments with $x(t)$ and $y(t)$.
$~$ Thus you are evaluating the partial differential of the field $f(u,v)$ over the $t$ parametised curve $\{\langle x(t),y(t)\rangle\}$
So the chain rule is actually:
$$\dfrac{\partial f(x(t),y(t))}{\partial t}=\left.\dfrac{\partial f(u,v)}{\partial u}\right\vert_{\raise{2ex}{u:=x(t)\\v:=y(t)}}\cdotp\dfrac{\partial x(t)}{\partial t}+\left.\dfrac{\partial f(u,v)}{\partial v}\right\vert_{\raise{2ex}{u:=x(t)\\v:=y(t)}}\cdotp\dfrac{\partial y(t)}{\partial t}$$
But this is annoying to read and write, so the shorthand is used for convenience.
Best Answer
$$c=\ln(e^{z_1}+e^{z_2})-z_y$$ $$\frac{\partial c}{\partial z_1}=\frac{e^{z_1}}{e^{z_1}+e^{z_2}}$$ $$\frac{\partial c}{\partial z_2}=\frac{e^{z_2}}{e^{z_1}+e^{z_2}}$$ remember with partial derivatives: it is the same as taking a derviative but you treat all other varables as constants so for each one we can say: $$c=\ln(e^{z_1}+a_1)-a_2$$ $$\frac{dc}{dz_1}=\frac{e^{z_1}}{e^{z_1}+a_1}$$ now substitute in your other variables as the "constants"
The mistake you made in each step was the following: $$\frac{\partial u_2}{\partial z_1}=\frac{\partial u_2}{\partial z_2}=0$$ Other than this your answer is correct :)