You looked at each of the three terms separately. You have $|x-1|<\frac{1}{10}$. You should have $|x+1|<\frac{21}{10}$ and you should have $\frac{1}{|x+3|}<\frac{10}{39}$. The critical point for your method is that you have to take the worst case each time.
The three inequalities then multiply together to give $\left|\frac{x^2-1}{x+3}\right|<\frac{21}{390}<\frac{1}{13}$.
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But note that this method only works because you were given a limit which is substantially larger than the largest possible value of the expression. If you had been asked to show that $$\left|\frac{x^2-1}{x+3}\right|<\frac{1}{19}$$ which is still true for $|x-1|<\frac{1}{10}$ the method would fail.
For a tougher limit the tool of choice is calculus, but one can often solve problems without it. For example in this case we could prove the $\frac{1}{19}$ limit by using the properties of quadratics.
Note that $x+3$ is positive throughout the range $0.9<x<1$, whereas $x^2-1<0$ for $0.9<x<1$ and $x^2-1>0$ for $1<x<1.1$ so we need to show that $19(1-x^2)<x+3$ for $0.9<x<1$ and $19(x^2-1)<x+3$ for $1<x<1.1$. We have
$$19(1-x^2)-(x+3)=-19x^2-x+16=-19\left(x+\frac{1}{38}\right)^2+16\frac{1}{76}$$ which is strictly decreasing for $x>\frac{1}{38}$. So its largest value in the range $0.9\le x\le 1.1$ is $19(1-0.9^2)-3.9=-0.29<0$ at $x=0.9$.
Similarly $$19(x^2-1)-(x+3)=19\left(x-\frac{1}{38}\right)^2-22\frac{1}{76}$$ which is strictly increasing in the range $1\le x\le 1.1$, so its largest value in the range is $19(1.21^2-1)-4.1=-0.11<0$ at $x=1.1$.