Counting Points on an Elliptic Curve over Finite Field $X^2Y+Y^2Z+Z^2X=kXYZ$ Let $p$ be a prime number(maybe generic large odd prime). Let $k$ be integer with $(k,p)=1$.
The equation in question is $$X^2Y+Y^2+X=kXY.$$
In homogeneous form, it is alternating
$$X^2Y+Y^2Z+Z^2X=kXYZ.$$
The elliptic curve always contain three points (0,0,1), (1,0,0) and (0,0,1). 
My question: Is there any formula relating the number of points in $\mathbb Z/p\mathbb Z$ to something else ?
$$\#\{(X,Y)\in (\mathbb Z/p\mathbb Z)^2 : X^2Y+Y^2Z+Z^2X=kXYZ\}= \;?$$
I am expecting the right-hand side be something related to certain characters either cubic or related to cubic fields (involving $k$ for sure).
 A: It's not clear to me why you expect $\#E(\mathbb F_p)$ to be related to cubic characters or to $\mathbb Q(\sqrt[3]{k})$. As far as I can tell, your curves are not cubic twists of a single curve. The easiest way to check this, as Kevin Casto suggests, is to covert to Weierstrass form and compute the $j$-invariant. (Added: Using Rene's calculation, 
$j(E_k)=\frac{k^{12}
 - 72 k^9
 + 1728 k^6
 - 13824 k^3}{k^3
 - 27}$.)
Thus you don't have a twist family. On the other hand, your question "Is there any formula relating the number of points in $\mathbb Z/p\mathbb Z$ to something else?" has an affirmative answer, but probably not  one you want. Namely if you fit $\#E_k(\mathbb F_p)$ appropriately into an $L$-series, then the Mellin transform will be a modular form (courtesy of Wiles et al.), so the number of points is related to the coefficients of the Fourier expansion of a modular form.
For $k=1$, let $N_p=\#E_1(\mathbb F_p)$. Then the first few values are
$$N_{5}=9, N_{7}=9, N_{11}=6, N_{13}=13, N_{17}=21, N_{19}=18, N_{23}=24, N_{29}=24, N_{31}=36, N_{37}=45, N_{41}=42, N_{43}=45, N_{47}=45.$$  This sequence is not in the OEIS. 
A: The number of projective solutions is $p+1$ plus the Legendre symbol $\left( \frac{k}{p}\right)$ plus a finite symbol hypergeometric sum involving $(k/3)^3$.
To show this, take Rene's Weierstrass form
$  w^2=z^3+k^2z^2+8kz+16$
Multiply z by $k^4$ and $w$ by $k^6$.
$  w^2=z^3+k^6z^2+8k^9z+16k^{12}$
We obtain a polynomial in $t=k^3$ with discriminant $256 t^8 (t-27)$ so it has three singular points $0,1,\infty$. Hence it has singularities only at $t=0,t=27, t=\infty$. Because the original curve has good reduction at $k=0$, the new curve has potentially good reduction, which because the discriminant has a zero of order $8$ means monodromy of order $3$. At $t=27$, there is semistable reduction, hence unipotent local monodromy. At $t=\infty$ there was semistable reduction for the original curve (easiest to tell from the original form $XYZ$) so the cubic pullback must also have semistable reduction, so the local monodromy is minus a unipotent.
By Katz's hypergeometric rigidity that means the $H^1$ of this family of curves is geometrically isomorphic to the $\mathcal H\begin{pmatrix} 1 ,1 \\ \chi_3 ,\overline{\chi}_3 \end{pmatrix}$ hypergeometric sheaf, so the trace function is a constant times the hypergeometric trace function
$\sum_{a,b \in \mathbb F_p | ab =t/27 } \chi_3\left(\frac{a}{a-1}\right) \overline{\chi}_3\left(\frac{b}{b-1}\right)  = \sum_{a,b \in \mathbb F_p | ab =27/t } \chi_3\left(a-1\right) \overline{\chi}_3\left(b-1\right)$
where $\chi_3$ is a character of order $3$
so the count of points on your original curve should have the form
$$p+1 + c_p  \sum_{a,b \in \mathbb F_p | ab =27/t} \chi_3\left(a-1\right) \overline{\chi}_3\left(b-1\right) $$
where $c_p$ is a constant (I think one can check a root of unity) depending on $p$.
Probably there is a direct and elementary proof of this by manipulating the sum, which should also give the constant.

If we consider the map from this curve to the curve of $A:B:C:D$ in $\mathbb P^3$ satisfying $A+B+C=kD$ and $ABC=D^3$ given by $A=X^2Y, B= Y^2Z,C=Z^2X, D=ABC$ we see that it has an inverse - note that $X/Z = A/D$, $Y/X= B/D$, so our inverse can take $(A:B:C:D)$ to $(AD, AB, D^2)$. So it's sufficient to count the projective solutions to this equation (unless there are some singularities and the inverse doesn't exactly exist - I don't want to calculate this right now but it would be easy to).
For an additive character $\psi$ of $\mathbb F_p$, the exponential sum
$\sum_{A,B,C,D | ABC = D^3} \psi(A+B+C-kD)$ 
is $p$ times the number of projective solutions of the equation minus the number of projective solutions of $ABC=D^3$ (which is $(p-1)^2$ where $D$ is nonzero plus $3p$ where $D$ is zero, or $(p^2+p-1)$) plus $1$.
Hence the count of points is $p+1$ plus $\frac{1}{p} \sum_{A,B,C,D | ABC = D^3} \psi(A+B+C-kD)$ 
Up to some other terms which I believe magically cancel, this is $\frac{1}{p(p-1) } \sum_{A,B,C,D ,\chi} \psi(A+B+C-kD)\chi(A) \chi(B) \chi(C) \chi^{-3}(D) $ which can be written in terms of Gauss sums as $\frac{1}{p(p-1) } \sum_\chi G(\chi,\psi)^3 G(\chi^{-3},\psi) \chi^3(-k)$.
If $p=1$ mod $3$ one can make cubic characters appear using a Hasse-Davenport relation $G(\chi^{-3},\psi) = - G(\chi^{-1},\psi) G(\chi_3 \chi^{-1},\psi) G(\overline{\chi}_3 \chi^{-1},\psi)\chi^{-3}(3)/(G(1,\psi) G(\chi_3,\psi) G(\overline{\chi}_3,\psi)$
using $G(\chi,\psi) G(\chi^{-1},\psi) = p \chi(-1)$ and multiplicativity of $\chi$ we obtain
$$\frac{1}{(p-1)(G(1,\psi) G(\chi_3,\psi) G(\overline{\chi}_3,\psi) } \sum_\chi G(\chi,\psi)^2 G(\chi_3 \chi^{-1},\psi) G(\overline{\chi}_3 \chi^{-1},\psi) \chi(k^3/27)$$
where the inner sum is another formulation of the hypergeometric sum  $\mathcal H\begin{pmatrix} 1 ,1 \\ \chi_3 ,\overline{\chi}_3 \end{pmatrix}$ applied to $\chi^3/27$.
