Determine that the function $f(x)=\sqrt{x^2-x-6} \text{ in } x_0=3$ is continous with the $\varepsilon$-$\delta$-definition of limit/criterion 
[Proof-verification] Determining whether the function $f(x)=\sqrt{x^2-x-6} \text{ in } x_0=3$; $x_0\in \mathbb{R}$ is continous $\color{red}{\text{ in  }x_0}$ or not with the
  $\varepsilon$-$\delta$-definition of limit/criterion:


$$f(x)=\sqrt{x^2-x-6} \text{ in } x_0=3$$
Proof:
Let $\varepsilon>0$ and ${\mid x-x_0\mid}<\delta \iff {\mid x-3\mid}<\delta$
\begin{align}
{\mid f(x)-f(x_0)\mid}&= {\mid\sqrt{x^2-x-6}-\sqrt{3^2-3-6}\mid}\\
& ={\mid\sqrt{x^2-x-6}\mid}\\
& ={\mid\sqrt{x+2} \cdot \sqrt{x-3}\mid}\\
& <\sqrt{x+2}\cdot \sqrt{x+2}\\
& = x+2\\
& < x-3+5 \\
& <\delta+5 =: \varepsilon \\
& \iff \delta = \varepsilon -5
\end{align}
$\implies$ the function is continous in $x_0=3 \qquad \qquad \qquad \qquad \qquad_\blacksquare $

Is this proof correct?

First of all, the function $f$ is not defined on $\mathbb{R}$ but for 
$$
x\in(-\infty,-2]\cup[3,+\infty).
$$ 
So one can only talk about its continuity of $f$ at $x=3$ from the right. 
For $x>3$, you are right to get
$$
|f(x)-f(3)|=\sqrt{x-3}\sqrt{x+2}.
$$
Note that you don't need the absolute value for the square root terms. 
But then you made a mistake: the $\delta$ you get must be positive. 
The term $\sqrt{x-3}$ should not be dropped and it would give you the desired $\delta$. 
Consider instead for $0<x-3<1$ the inequality
$$
\sqrt{x-3}\sqrt{x+2}\leq 6\sqrt{x-3}\le\varepsilon.
$$
