Question: Let $v:[0,\infty)\rightarrow \mathbb{R}^n$ be a function such that $\forall t\ge 0$, $v\_i(t)\ge 0$ and
$$ ||v(t)||\le \beta ||v(0)||e^{-at}, t\ge 0$$
with $\beta,a>0$
can I conclude that for all $i=1,...,n$ there exist $\xi\_i,\eta\_i>0$ s.t. $$v\_i(t) \le \xi\_iv\_i(0)e^{-\eta\_it}$$?

Answer: 
Your solution is correct but your notation is inconsistent.

$\frac{dy}{dx}$ indicates the derivative of the function $y$ with respect to $x$; you can't write $\frac{dy}{dx}[x]$ as the latter would mean the derivative of $y$ times $x$.

You can either use $\frac d{dx} x$ (and so also $\frac d{dx} \ln x$) or simply $(\ln x)'$.

Note also that your original function is $f(x)$, so you're computing $\frac {d}{dx} f(x) = f'(x)$; so at the end you can't write $y'$ because $y$ was not defined in the first place
Your answer is correct but your notation inconsistent, as has been pointed out by other users. Another way of seeing that your answer is correct is to integrate $\ln x$.

$$\int \ln x \, \mathrm{d}x \stackrel{\text{IBP}}{=} x \ln x - \int \frac{x}{x} \, \mathrm{d}x = x \ln x - x + \mathrm{C}$$

Where the integration by parts is done by setting $u = \ln x$ and $\mathrm{d}v = 1$, a sneaky trick. Then $v = x$ and $x \, \mathrm{d}u = \mathrm{d}x$. The result follows. 

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I saw that you were looking for a way to find the derivative using logarithmic differentiation, this is done below: 

$$\ln f(x) = \ln \left(x \ln x - x\right) = \ln \left(x(\ln x - 1)\right) = \ln x + \ln (\ln x - 1).$$

So differentiating implicitly with respect to $x$ yields $$\frac{f'(x)}{f(x)} = \frac{1}{x} + \frac{\frac{1}{x}}{\ln x - 1} = \frac{1}{x} + \frac{1}{x(\ln x -1)}$$

Multiplying through by $f(x)$ yields 

$$f'(x) = \frac{x(\ln x - 1)}{x} + \frac{x(\ln x -1)}{x(\ln x - 1)} = \ln x -1 + 1$$

So we have $$f'(x) = \ln x.$$