It should be
$$ e^{-2t}(-\sin(4t) \cdot 4)+ \cos(4t)(e^{-2t}(-2)).$$
Your logic is right, but
$$\frac{d}{dt}(\cos(4t)) = -4\sin(4t),$$
not $-4\sin(4t)\cos(4t).$
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:
$$\begin{align}t&=9\\s&=3\\u&=4\\y&=2\\&\dots\\\frac{dy}{dt}&=-\frac23\end{align}$$
Best Answer
For the exponential function, $$\frac{d}{dx}\left(e^{f(x)}\right) = e^{f(x)}f'(x).$$
Here, $e^{f(x)} = e^{3\sqrt x}$, so $f(x) = 3\sqrt x = 3x^{1/2}$. So then we must have $$\frac{d}{dx}\left(e^{3\sqrt{x}}\right) = e^{3\sqrt{x}}\left(\frac{3}{2}x^{-1/2}\right) = \frac{3e^{3\sqrt x}}{2\sqrt x}$$
As per comments: Yes, $e^x$ is unique in comparison to $x^n$, in many ways, including the fact that the first is a very distinguished constant raised to a variable power whereas the second is a variable raised to a constant. So the power rule does not apply to $e^x$, nor does it apply to any constant raised to a variable. And with the unique constant $e$: recall, $\;\frac{d}{dx}(e^x) = e^x$.