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: 
The answer is correct, but your notation is a bit off. Instead, write

\begin{align}
f(x) &= x \ln (x) - x \\
\frac{\mathrm{d}}{\mathrm{d}x}f(x) &= \frac{\mathrm{d}}{\mathrm{d}x} (x \ln (x) - x) = x \frac{\mathrm{d}}{\mathrm{d}x} \ln(x) + \ln(x) \frac{\mathrm{d}}{\mathrm{d}x} x - \frac{\mathrm{d}}{\mathrm{d}x} x = \cdots
\end{align}

Note the difference between $\frac{\mathrm{d}y}{\mathrm{d}x}$ in your case (incorrect) and $\frac{\mathrm{d}}{\mathrm{d}x}$ above (correct).
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 $$f'(x) = \ln x.$$