Polynomial $x^3- xy^3$ and the like over finite fields. Let $f_{a,n}(x_1,x_2)$ be a polynomial in $\mathbb{F}_p[x_1,x_2]$, where $\mathbb{F}_p$ is a finite field or oder $p$ (perhaps, we may first assume that $p$ is prime) depending on $a\in\mathbb{F}_p$ and an integer $1\le n\le p-1$. Let $f_{a,n}$ be defined as follows as $f_{a,n}(x_1,x_2) := ax_1^n - x_1 x^n_2$. I am trying to investigate the following question: for what $a$ and $n$ is the polynomial $f_{a,n}$ surjective as a function from $\mathbb{F}_p\times\mathbb{F}_p$ to $\mathbb{F}_p$.
As the question seems rather difficult, let us first concentrate on the case when $n = 3$, $a = 1$. It appears that the polynomial $f = f_{1,3} = x_1^3-x_1 x_2^3$ is surjective over any field of odd (not necessarily prime) characteristics $\mathbb{F}_p$, $3\le p\le 1000$, $p\neq 19$. I would like to understand the number theory behind this result. Why is the case $p = 19$ special or 'exceptional'? It is interesting to note that for all other values of $3 \le p \le 1000$, $p$ odd, the polynomial $f_{a,3}=a x_1^3-x_1 x_2^3$ is surjective for any $a\in\mathbb{F}_p$! (this is not $p!$, it is me being astinoshed by this result).
For other combinations of $a$ and $n$ there are also such 'exceptional' values of $p$. 
Does anyone know if polynomials of the type $x^3-xy^3$ have been studied over finite fields?
 A: I'm gonna admit defeat, reach for a bigger tool, and supplement Thomas Andrews' answer by swatting this fly with...
We are interested in solvability of the equation
$$
x^3-xy^3=a,\qquad{(*)}
$$
where $a\in\mathbb{F}_p$ is a constant, and $x,y$ range over the same field. It is not a restriction to assume that $p>3$ and $a\neq0$. The corresponding projective curve is the zero locus of the homogeneous polynomial
$$
F(X,Y,Z)=X^3Z-XY^3-aZ^4.
$$
Looking at the partial derivatives $\partial F/\partial X$, $\partial F/\partial Y$, $\partial F/\partial Z$ we see that they vanish simultaneously only when $X=Y=Z=0$, so the curve has no singularities. Furthermore, it has exactly two points on the line at infinity $Z=0$, namely $[1:0:0]$ and $[0:1:0]$. 
The genus of a non-singular plane curve given by a degree $n$ equation is $g=(n-1)(n-2)/2$, so the genus of our curve is $g=(4-1)(4-2)/2=3$. The Hasse - Weil bound says that the number of $\mathbb{F}_p$-rational points $N_p(C)$ on a smooth model of a curve $C$ of genus $g$ satisfies the inequality
$$
|N_p(C)-(p+1)|\le 2g\sqrt{p}.
$$ 
This gives us a lower bound $N_p(C)\ge p+1-6\sqrt{p}$ for our curve.
Let $N(a)$ be the number of solutions of $(*)$. Counting for the two points at infinity we end up with a lower bound
$$
N(a)\ge p-1-6\sqrt{p}.
$$
We see that this lower bound is positive for all primes $p>37$, so if you have checked the primes up to $37$, and found out that $p=19$ is the only problematic case, you can stop looking.
A: As noted above, $f_{a,3}$ is onto when $p\not\equiv 1\pmod 9$. If $p\not\equiv 1\pmod 3$ then ever element of $\mathbb F_p$ is a perfect cube, so we can find a solution with $x_2=0$.
When $p\equiv 1\pmod 3$ and $p\not\equiv 1\pmod 9$, we get a primitive cube root of unity, $\zeta^3=1$ in $\mathbb F_p$, and $\zeta$ itself is not a cube.
Then for any $z\in\mathbb F_p$, one of the following is a perfect cube: $$z-a,\frac{z-a}{\zeta},\frac{z-a}{\zeta^2}$$ 
In which case we can pick $x=1,x=\zeta,$ or $x=\zeta^2$, respectively and find a solution.
In general, $z\neq 0$ is in the image of $f_{a,3}$ if and only if $\frac{z-ax^3}{x}$ is a cube for some $x\neq 0$. If it was purely random whether that value was a cube, it would be highly unlikely that there were any such $z$ when $p$ is large. That's not a rigorous argument, just a heuristic one.
So I suspect that the reason $19$ is a counter-example is that $19\equiv 1\pmod 9$ and $19$ is small.
One other thing to note is that if $z$ is not in the image and $w\in\mathbb F_p^\times$ then $w^9z$ is also not in the image. Because if $w^9z=ax^3-y^3x$ then $$z=a(xw^{-3})^2-(yw^{-2})^3(xw^{-3})$$
This means that for any $p\equiv 1\pmod 9$ we only actually have to check $6$ values for $z$ (because three of the nine equivalence classes in $\mathbb F_p^\times/(\mathbb F_p^\times)^9$ are cubes, and we know those are okay.)
