Prove: if $|x-1|<\frac{1}{10}$ so $\frac{|x^2-1|}{|x+3|}<\frac{1}{13}$ 
Prove: $$|x-1|<\frac{1}{10} \rightarrow \frac{|x^2-1|}{|x+3|}<\frac{1}{13}$$

$$|x-1|<\frac{1}{10}$$
$$ -\frac{1}{10}<x-1<\frac{1}{10}$$ 
$$ \frac{19}{10}<x+1<\frac{21}{10}$$
$$|x+1|<\frac{19}{10}$$
Adding 4 to both sides of $$ -\frac{1}{10}<x-1<\frac{1}{10}$$ gives:
$$\frac{39}{10}<x+3<\frac{41}{10}$$
$$|x+3|<\frac{39}{10}$$
Plugging those results in $$\frac{|x-1|*|x+1|}{|x+3|}<\frac{1}{13}$$
We get: $$\frac{\frac{1}{10}*\frac{19}{10}}{\frac{39}{10}}<\frac{1}{13}$$
$$\frac{19}{390}<\frac{1}{13}$$ Which is true, is this proof is valid as I took the smallest intervals, like $|x+3|<\frac{39}{10}$ and not $|x+3|<\frac{41}{10}$?

Your method (what you are trying to do) is correct, but you have a few errors.

$$ \frac{19}{10}<x+1<\frac{21}{10}$$
$$|x+1|<\frac{19}{10}$$

This is wrong. It should be
$$|x+1|\lt \frac{21}{10}$$

$$\frac{39}{10}<x+3<\frac{41}{10}$$
$$|x+3|<\frac{39}{10}$$

This is wrong. It should be
$$|x+3|\lt\frac{41}{10}$$
but we use
$$|x+3|\color{red}{\gt} \frac{39}{10}\iff \frac{1}{|x+3|}\lt \frac{10}{39}$$
since $|x+3|$ is in the denominator.
Therefore, we get
$$\frac{|x^2-1|}{|x+3|}\lt\frac{1}{10}\cdot\frac{21}{10}\cdot\frac{10}{39}=\frac{7}{130}\lt \frac{10}{130}=\frac{1}{13}$$
