If I roll three dice at the same time, how many ways the sides can sum up to $13$? If I rolled $3$ dice how many combinations are there that result in sum of dots appeared on those dice be $13$?

It's the coefficient of $x^{13}$ in the product $(x+x^2 + x^3 + x^4+ x^5 + x^6)^3$.
To see this, note that to compute that coefficient we have to identify all ways we can form $x^{13}$ by picking one term from each of the three terms $(x + x^2 + x^3 + x^4 + x^5 + x^6)$ we have; we could have $x$ from the first, $x^6$ from the second and the third and this would correspond to throwing $(1,6,6)$ with the three different dice (which we imagine to have different colours to distinguish them). This choice gives us one way to get $x^{13}$ in the final gathering of terms, and all other choices (so pairs $(a,b,c)$ with $a + b + c = 13, 1 \le a,b,c \le 6$) give us one extra power of $x^{13}$. So the final coefficient just counts all those triples. 
E.g. try this with two dice: $$(x+x^2 + x^3 + x^4+ x^5 + x^6)^2 = x^2 + 2x^3 + 3x^4 + 4x^5 + 5x^6 + 6x^7 + 5x^8 + 4x^9 + \\ 3x^{10} + 2x^{11} + x^{12}$$ and we see that the coefficient of $x^n$ is just the number of ways we can throw $n$ with two dice.
Write this as $(x(1+x+x^2+x^3+x^4+x^4))^3 = x^3(1+ x + x^2 + x^3 + x^4 + x^5)^3$, so we are looking for the coefficient of $x^{10}$ in $(1+ x + x^2 + x^3 + x^4 + x^5)^3$. 
The fancy way to do this is to write $(1+x+x^2+\ldots+x^5) = \frac{1-x^6}{1-x}$ (standard geometric series) and so $(1+ x + x^2 + x^3 + x^4 + x^5)^3 = (1-x^6)^3 (1-x)^{-3}$.
The first term on the right can be evaluated using the binomial formula as $1 - 3x^6 + 3x^{12} - x^{18}$.
The second term on the right can be evaluated by the generalised binomial formula as $\sum_{k=0}^{\infty} {k+2 \choose k} x^k$.
So to get $x^{10}$ from the product of these, we get the $1$ from the first times the ${12 \choose 10}x^{10}$ from the second and the $-3x^6$ from the first times ${6 \choose 4}x^4$ of the second. Other terms have too high powers of $x$.
So the answer is ${12 \choose 10} - 3{6 \choose 4} = 21$.
You can use Wolfram alpha to expand the original polynomial $(1+x+\ldots+x^5)^3$ and we get $$x^{15}+3 x^{14}+6 x^{13}+10 x^{12}+15 x^{11}+21 x^{10}+\\25 x^9+27 x^8+27 x^7+25 x^6+21 x^5+15 x^4+10 x^3+6 x^2+3 x+1$$
The alternative is simple enumeration. But I like the complicated ways, as they generalise to more dice and higher sums. E.g. in the final expansion we see that there are $25$ ways to throw $9+3 = 12$ with three dice, etc. We get all the probabilities for all the sums at the same time.
