What is the correct answer for $\mathcal{L}^{-1}\left(\frac{p^2}{(p-3)^2}\right)?$ What is the correct answer for $$\mathcal{L}^{-1}\left(\frac{p^2}{(p-3)^2}\right)?$$
Using partial fraction technique I got the answer as: $\delta(t)+e^{3t}9t+6e^{3t}$ and uing shifting thorem for inverse Laplace I gotthe answer as: $e^{3t}(\delta(t)+6+9t)$.
 A: The frequency shift property of Laplace Transform states that:
$$
\bbox[5px,border:1.1px solid black] {
F(p+a)\iff e^{-at}\cdot f(t)
}
$$
It's implicit in this property that you must apply the inverse Laplace transform to $F(p)$, and then multiply the result by $e^{-at}$. 

We can use it in the case you've presented, look:
$$
F(p-3)=\frac{p^2}{(p-3)^2}
$$
If $p'=p-3$, then $p=p'+3$. So, we have:
$$
F(p'+3-3)=\frac{(p'+3)^2}{(p'+3-3)^2}\iff F(p')=\frac{(p'+3)^2}{(p')^2} \implies F(p)=\frac{(p+3)^2}{p^2}
$$
Applying the inverse Laplace transform to $F(p)$, we have:
$$
\begin{alignat}{1}
\mathscr{L}^{-1}\left[F(p)\right]&=\mathscr{L}^{-1}\left[\frac{(p+3)^2}{p^2}\right]
\\&=\mathscr{L}^{-1}\left[\frac{p^2+6p+9}{p^2}\right]
\\&=\mathscr{L}^{-1}\left[1+\frac 6 p + \frac{9}{p^2}\right]
\\&=\delta(t)+6+9t=f(t)
\end{alignat}
$$
Using the frequency shift property:
$$
\bbox[5px,border:1.1px solid black] {
\mathscr{L}^{-1}[F(p-3)]=e^{-(-3)t}\cdot f(t)=e^{3t}\cdot[\delta(t)+6+9t]
}
$$
So, both answers are correct

Having that in mind, we need to prove that $$e^{3t}\delta(t)=\delta(t)$$
First, it's necessary to make some clarifications:

A "delta function" is not really a function, it's what is called a "generalized function", or distribution.  So, when you see people write $\delta(0) = \infty$, this is just a "formality", and a very misleading one at that.  The only correct way to think about a delta function is under an integral sign with another ("normal") function $f(x)$:
$$
\int \delta(x)f(x)dx
$$ 
  This expression makes perfect sense, and as long as $f(x)$ is "nice enough" (e.g. continuous at 0), it will evaluate to $f(0)$.
                                                                                                          Source: this answer, given by the user icurays1 

So, all we need is this definition: 
$$
\bbox[9px,border:1.1px solid black] {
\int f(t)\cdot \delta(t)\,\,dt \triangleq f(0)
}
$$
The proof:
$$
\begin{align}
\delta(t)=1\cdot\delta(t) &\implies \int 1\cdot \delta(t)\,\,dt=1
\\ e^{3t}\cdot\delta(t)&\implies \int e^{3t}\cdot\delta(t)\,\,dt=e^{3\cdot\color{red}{0}}=1
\end{align}
$$
So, we can conclude that:
$$
\\ \int\delta(t)\,\,dt=\int e^{3t}\cdot\delta(t)\,\,dt \implies\delta(t)=e^{3t}\delta(t)
$$

One important note:
You should avoid interpreting the Dirac Delta Function this way:
$$
\delta(x) = \left\{\begin{array}{cc}
\infty & x = 0 \\
0 & x\neq 0 
\end{array}\right.
$$
I'll show one example that illustrates why: 
What is $\mathscr{L}\left[\,\delta(t)\,\right]$? And $\mathscr{L}\left[\,\delta(3t)\,\right]$?
Using the above interpretation, we have:
$$
\begin{alignat}{0}
\delta(t) = \left\{\begin{array}{cc}
\infty & t = 0 \\
0 & t\neq 0 
\end{array}\right.
&&\delta(3t) = \left\{\begin{array}{cc}
\infty & 3t = 0 \\
0 & 3t\neq 0 
\end{array}\right.
= \left\{\begin{array}{cc}
\infty & t = 0 \\
0 & t\neq 0 
\end{array}\right.
=\delta(t)
\end{alignat}
$$
So, since we have the same "function"... $\mathscr{L}\left[\,\delta(t)\,\right]$ must be equal to $\mathscr{L}\left[\,\delta(3t)\,\right]$, right? WRONG!
$$
\begin{alignat}{1}
\mathscr{L}\left[\,\delta(t)\,\right]&=1 \\
\mathscr{L}\left[\,\delta(3t)\,\right]&=\frac 1 3
\end{alignat}
$$
But Laplace Transform Uniqueness Theorem states that:

$$\mathscr{L}[\,f_1\,]\neq\mathscr{L}[\,f_2\,] \iff f_1 \neq f_2$$

(And this theorem is valid for Dirac Delta 'function').
Then:
$$
\mathscr{L}\left[\,\delta(t)\,\right] \neq \mathscr{L}[\,\delta(3t)\,] \iff \delta(t) \neq \delta(3t)
$$

Note: 
One could argue: "But Dirac Delta 'function' isn't continuous. You have omitted the part of Uniqueness theorem that says that $f_1$ and $f_2$ must be continuous on $[0,∞)$.Then, the theorem is not valid for this case." 
My counter argument: Yes, Delta 'function' isn't continuous. But there is a theorem stronger than this one which says that Uniqueness also applies to the Dirac Delta Function. See this excerpt, extracted from Laplace Transform Lecture Notes - Ante Mimica (Suggested by Calvin Khor here)
A: Both are correct as $\delta(t)=e^{3t}\delta(t)$. See the definitions of delta function and delta sequence for more details.
A: Both are correct because $\delta(t)=e^{3t}\delta(t).$
$$\delta(t)=\{_{0,\;\;x\neq0}^{+\infty,\;x=0}$$
and
$$\int_{-\infty}^{+\infty}\delta(t)dt=1$$
Likewise
$$e^{3t}\delta(t)=\{_{0,\;\;x\neq0}^{+\infty,\;x=0}$$
and
$$\int_{-\infty}^{+\infty}e^{3t}\delta(t)dt=1$$
since $$\int_{-\infty}^{+\infty}f(t)\delta(t)dt=f(0)$$
Therefore
$$\delta(t)=e^{3t}\delta(t)$$
