Which of these functions extend by continuity?

1. As $x\to0$, $\frac1{x^2}\to+\infty$ so $-\frac1{x^2}\to-\infty$ and $e^{-1/x^2}\to0$. The limit is $0$, so $f_1$ extends continuously with $f_1(0)=0$.

2. The two sides behave differently.
- For $x\to0^+$, $-\frac1x\to-\infty$ and $e^{-1/x}\to0$.
- For $x\to0^-$, $-\frac1x\to+\infty$ and $e^{-1/x}\to+\infty$.

There is no finite limit at $0$, so $f_2$ cannot be extended.

3. Put $t=x+1\to0$. Then $f_3=\sin t\,\ln|t|=\dfrac{\sin t}{t}\cdot t\ln|t|$. We know $\frac{\sin t}t\to1$ and $t\ln|t|\to0$, so the limit is $0$. Hence $f_3$ extends continuously with value $0$ at $-1$.

Remark. Always test both sides: a function like $e^{-1/x}$ is perfectly well behaved on the right of $0$ yet explodes on the left.