I thought delta function's three properties.
$\delta(0)=\infty$, $\displaystyle\int\delta(t)dt=1$, and $\delta(t)=\delta(-t)$.
Therefore, if I do scaling by $a$, integral's value will be $1/|a|$.
$$ \therefore\delta(at)=\frac1{|a|}\delta(t) $$
In turn, differentiate both parts with t.
$$ \frac{d}{dt}\bigl[\delta(at)\bigr]=\frac{d}{dt}\left[\frac1{|a|}\delta(t)\right] $$
$$ \Leftrightarrow\ \ \ \delta'(at)=\frac1{|a|}\delta'(t) $$
Is this correct?
No: $\frac{d}{dt}(\delta(at)) = a \delta'(at)$. The chain rule (suitably interpreted) still applies.
Whether to consider it correct or not after this modification, depends on how formal you want to be in your treatment of distributions. (And since you say things like $\delta(0) = \infty$, I'm guessing you don't want to be very formal.)