People don't normally write this as the chain rule, but in reality it is:
$$\frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}$$
Now let's see what you have. You found the derivatives correctly:
$$\begin{align}y' &= -\frac{3}{(u-1)^2}\\
u' &= 6(3s-7)\\
s' &= \frac{1}{2\sqrt{t}}\end{align}$$
Because the chain rule is algebraic, it's easier to see if I write the derivatives in the more technical notation to visualize what you took the derivative against (i.e. $y'=x\Rightarrow \frac{dy}{dx}=x$):
$$\begin{align}\frac{dy}{du} &= -\frac{3}{(u-1)^2}\\
\frac{du}{ds} &= 6(3s-7)\\
\frac{ds}{dt} &= \frac{1}{2\sqrt{t}}\end{align}$$
How would you get to $\frac{dy}{dt}$? Well, you'd use the above and do a little dimensional analysis. First start with the $\frac{dy}{du}$. Now you need to cancel the $du$, so multiply by $\frac{du}{ds}$. Need to cancel the $ds$ too, so multiply by $\frac{ds}{dt}$. Now you are left with $\frac{dy}{dt}$ and it was pretty easy. So again, we have:
$$\frac{dy}{dt}=\frac{dy}{du}\frac{du}{ds}\frac{ds}{dt}=\left(-\frac{3}{(u-1)^2}\right)\left(6(3s-7)\right)\left(\frac{1}{2\sqrt{t}}\right)$$
Now you need to evaluate it at $t=9$. Plug $9$ into the original $s$ function to find $s$. Using $s$, find $u$ from its function. Likewise, find $y$ using $u$ and you've got all the variables you need.
If my math is correct, you have:
$$\begin{align}t&=9\\s&=3\\u&=4\\y&=2\\&\dots\\\frac{dy}{dt}&=-\frac23\end{align}$$
I will do out the example you mention in this third version of your question.
$P$ is a function $P:\mathbb{R}^6\to\mathbb{R}^2$ defined by
$$P\begin{pmatrix}
x\\y\\a\\b\\c\\d
\end{pmatrix} =\begin{pmatrix}a&b\\c&d\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}
ax+by\\cx+dy
\end{pmatrix}$$
$L$ is a function $:\mathbb{R}^3\to\mathbb{R}^2$ defined by
$$L\begin{pmatrix}
\kappa\\x\\y
\end{pmatrix}=\begin{pmatrix}
x^2\kappa\\xy\kappa
\end{pmatrix}$$
Therefore $L$ and $P$ cannot be composed. However, for any given value of $\kappa$ you can form the function $L_\kappa:\mathbb{R}^2\to\mathbb{R}^2$; for example if $\kappa=3$, then the function $L_3$ is defined by
$$L_3\begin{pmatrix}
x\\y\end{pmatrix}=
\begin{pmatrix}
3x^2\\ 3xy
\end{pmatrix}$$
Then $P$ can be composed with any choice of these functions to form $(L_\kappa\circ P):\mathbb{R}^6\to\mathbb{R}^2$.
Here are the derivatives: (I am using the $0$ subscripts to encourage you to think of them all as fixed numbers.)
$$P'\begin{pmatrix}
x_0\\y_0\\a_0\\b_0\\c_0\\d_0
\end{pmatrix} =\begin{pmatrix}
a_0 & b_0 & x_0 & y_0 & 0 & 0\\
c_0 & d_0 & 0 & 0 & x_0 & y_0
\end{pmatrix}$$
For any given $\kappa_0$,
$$L_{\kappa_0}'\begin{pmatrix}
x_0\\ y_0
\end{pmatrix}=\begin{pmatrix}
2\kappa_0x_0 & 0\\
\kappa_0y_0 & \kappa_0x_0
\end{pmatrix}$$
Therefore
$$\begin{align*}
(L_{\kappa_0}\circ P)'\begin{pmatrix}
x_0\\y_0\\a_0\\b_0\\c_0\\d_0
\end{pmatrix}&=L_{\kappa_0}'\left(P\begin{pmatrix}
x_0\\y_0\\a_0\\b_0\\c_0\\d_0
\end{pmatrix}\right)\cdot P'\begin{pmatrix}
x_0\\y_0\\a_0\\b_0\\c_0\\d_0
\end{pmatrix}\\\\\\
&=L_{\kappa_0}'\begin{pmatrix}
a_0x_0+b_0y_0\\c_0x_0+d_0y_0
\end{pmatrix}\cdot P'\begin{pmatrix}
x_0\\y_0\\a_0\\b_0\\c_0\\d_0
\end{pmatrix}\\\\\\
&=\underbrace{\begin{pmatrix}
2\kappa_0(a_0x_0+b_0y_0) & 0\\
\kappa_0(c_0x_0+d_0y_0) & \kappa_0(a_0x_0+b_0y_0)
\end{pmatrix}}_{\text{a }2\times 2\text{ matrix}}\underbrace{\begin{pmatrix}
a_0 & b_0 & x_0 & y_0 & 0 & 0\\
c_0 & d_0 & 0 & 0 & x_0 & y_0
\end{pmatrix}}_{\text{a }2\times 6\text{ matrix}}
\end{align*}$$
Best Answer
Let's recap some things: For general $f=(f_1,\ldots,f_m):U\subseteq\mathbb{R}^n\to\mathbb{R}^m$ (where $U$, the domain of $f$, is open in $\mathbb{R}^n$) and $x\in U$, $Df(x)$ is the Jacobian (or rather the linear transformation associated to it), given by $$Df(x)=\left[\frac{\partial f_i}{\partial x_j}(x)\right]_{\substack{i=1,\ldots,m\\j=1,\ldots,n}}$$ (wherever this makes sense). From this definition, the following are obvious:
When $m=1$, $Df(x)$ is simply a row, and in fact it is equal to $\nabla f(x)$, the gradient of $f$ at $x$.
When $n=1$, $Df(x)$ is a column, where each entry is simply the derivative of $f_j$ at $x$.
When $m=n=1$, $Df(x)$ is a number, equal to $f'(x)$.
Now recall the Chain Rule:
Now to your problem: to use the chain rule, you have to see $f$ as a composition of two simpler functions, which are differentiable everywhere. If we put $h:\mathbb{R}^n\to(0,\infty)$ as $h(x)=\Vert x\Vert^2$, and $k:(0,\infty)\to\mathbb{R}$ by $k(t)=\frac{t^2}{1+t}$, we have $f=k\circ h$, and it is easier to calculate $Dh$ and $Dk=k'$.
I'll leave the details to you, but here's what you should get: For $x=(x_1,\ldots,x_n)$, use the definition of $Dh(x)$ to find $Dh(x)=2[x_1\ x_2\cdots x_n]$.
For $k$, one-variable calculus gives $k'(t)=\frac{t^2+2t}{(1+t)^2}$.
By the chain rule, $$Df(x_1,\ldots,x_n)=2\frac{(\Vert x\Vert^4+2\Vert x\Vert^2)}{(1+\Vert x\Vert^2)^2}[x_1\cdots x_n]$$