Skewness of a random variable that have zero variance and zero third central moment If I have a random variable $x$, and the only information I know about it are:
$$ m_1=E[x]=c, \mu_2=var(x)=0, \mu_3=E[(x-m_1)^3]=0$$
Can I conclude that the function distribution is symmetric about c?
Symmetric distributions have skewness = 0, and the definition of skewness in wikipedia is:
$$ \gamma = \frac{\mu_3}{\sigma^3}$$
but in this case it is zero divided by zero!, but I can model this distribution by Dirac Impulse distribution since the variance is zero and the mean is c $f_x(x)=\delta(x-c)$ which is symmetric about c. But I'm not sure if I can say that the skewness is zero based on these information.
 A: Your title and some of the body of your post aren't consistent.
Your title (and last sentence for example) is about skewness. I take that to be the central issue.
However, one section in the body of your post asks about symmetry and makes the (incorrect) assertion that:  

Symmetric distributions have skewness = 0

This is not true in general! It's quite possible to have symmetric distributions that don't have zero third moment. (There are also asymmetric distributions with zero third moment. Moment-skewness and symmetry aren't as closely related as we might wish them to be.)
You ask about the moment-skewness coefficient of a point mass (in the title and in the later part of the question). [If your real question is "are point masses symmetric", then 'yes' is a perfectly sensible, (almost) trivial answer, from several approaches. But I assume the title Q is the main Q.]
Note that $0/0$ is undefined. Let me explain why "undefined" is the right answer for the skewness in this case.
When you have a ratio where the numerator and denominator are both 0, it's sometimes possible to consider the limit of the ratio (that is, to set up a sequence $a_i/b_i$ where $a_i$ and $b_i$ both go to $0$, and see what happens to the ratio as $i\to \infty$), and so perhaps (if the $a$'s and $b$s must be connected in a way that gives it one) obtain a value for the limit of the ratio that would be a reasonable answer to the title question.
But no such limit exists in this case.
Let's consider a sub-family of shifted-Gamma distributions with fixed shape parameter $\alpha$ and varying scale parameter, $\beta$, with the shift, $\theta$, chosen so that $\theta=c-\alpha\beta$ for some constant $c$. 
Plainly the expectation of each distribution is $c$.
The standardized-third-moment skewness coefficient of a gamma distribution (and hence of a shifted gamma) is $\frac{2}{\sqrt\alpha}$. 
Since we hold $\alpha$ fixed, the skewness of every member of the family is the same. 
Now consider what happens as the scale parameter, $\beta\to 0$. In the limit, the distribution is a point mass at $c$, but at every point along the way, the skewness is $\frac{2}{\sqrt\alpha}$.
$\hspace{0.7 cm}$ 
$^{\text{   The first few members of a sequence of distribution functions where expectation }=4,\text{ variance }\to 0 \text{ but } \gamma=\sqrt{2}}$
Hence, in the limit, the skewness for the point-mass distribution is also $\frac{2}{\sqrt\alpha}$.
But $\alpha$ could have been chosen to be any value at all, $>0$. Hence, every positive value for skewness is potentially available as a limit for the skewness at the point-mass.
Now consider that we could also flip this family about $c$ and have another family with negative skewness.
Hence, every negative value for skewness is potentially available as a limit for the skewness at the point-mass.
By similar argument, I can construct a symmetric family whose third moment is 0 (say a Gaussian family) with the same point mass limit.
Hence, $0$ is available as a limit for the skewness at the point-mass.
Similarly I can also obtain $\infty$, $-\infty$ and "undefined" (by taking a family with the relevant skewness coefficient and taking the limit in similar fashion).
As a result, every conceivable value (including no value at all) of the skewness coefficient is consistent with a point mass, simply by taking the limit of a different sequence of distributions.
So the right answer to the title question is "it's undefined", just as $0/0$ would have suggested to us.
[An argument from symmetry of the cdf of the point mass is irrelevant to the title question, which wasn't about symmetry, but about third moment skewness. If you think symmetry speaks to that question, note that the Cauchy distribution is symmetric, but its skewness isn't 0.]
A: This is a degenerate distribution with all of the mass at the mean=c, so yes it is symmetric.  
