Solutions of $x^2 -xy + y^2 \equiv 0 \pmod{43}$ 
$x^2 -xy + y^2 \equiv 0 \pmod{43}$

We could transform this into

$\frac{x^3 + y^3}{x+y} \equiv 0 \pmod{43}$

and thus

$x^3 + y^3 \equiv 0 \pmod{43}$

From what I could manage so far, I know that if $x = 3k$ , $y= 3k+1$, as $43 = 3k+1$.
Also there are no solutions for $43$ itself, $84$ and $127$.
How should I proceed?
 A: We have $x \equiv -y$ iff $x \equiv 0$ iff $y \equiv 0$.
Otherwise, as you have argued, $x^2 - xy + y^2 \equiv 0$ iff $x^3 + y^3 \equiv 0$ and this happens iff $(xy^{-1})^3 \equiv -1$.
So the solutions are $x=yu$, where $u$ is an element of order $6$ mod $43$, that is, $u=7$ or $u=37$.
A: The polynomial $p(T)=T^2-T+1$ is the sixth cyclotomic polynomial. Meaning that its zeros have order six.
Because $43-1=6\cdot7$ an element $a$ of the cyclic group $\Bbb{Z}_{43}^*$ is of order six if and only if it is a seventh power but is not a 14th or a 21st power.
Let's just list seventh powers and do a bit of screening:
$2^7=128\equiv-1$, no good. $3^7\equiv37$, this works because its square is $37^2\equiv-7$ and its cube is $\equiv-1$ so it has order six. Therefore $37$ is one zero of $p(T)$. The sum of the zeros is, by Vieta, equal to $1$, so the other zero is $1-37=7$.
Therefore 
$$p(T)=(T-7)(T-37).$$
This settles your question. Whenever $x=ay$ we have 
$$x^2-xy+y^2=y^2(a^2-a+1)=y^2p(a).$$
The solutions are thus $x=7y$ and $x=37y$, $y$ can be anything.
These are all the solutions because if we write $a=x/y$ the above calculation shows that $p(a)=0$. If $y=0$ then the only solution is obviously $x=0$.
A: If we were working in $\mathbb{Q}$ we could transform your equation into:
$$x^2 - xy + y^2 = 0 \qquad \iff \qquad \left(x- \frac{1}{2}y\right)^2 + \frac{3}{4}y^2 = 0.$$
Fortunately, both $2$ and $4$ are invertible modulo $43$:
$$(x - (2^{-1})y)^2 + 3(4)^{-1}y^2 \equiv 0 \pmod{43} \quad \iff \quad (x+21y)^2 + 33y^2 \equiv 0 \pmod{43}.$$
Let $z = x + 21y$. We have the equation $z^2 + 33y^2 \equiv 0 \pmod {43}$, or better yet,
$$z^2 \equiv 10y^2 \pmod{43}.$$
Now $10$ is a quadratic residue modulo $43$, since $15^2 \equiv 10 \pmod{43}$, so
$$z^2 \equiv (15y)^2 \pmod{43}$$
which means
$$z \equiv \pm 15y\pmod{43} \qquad \iff \qquad x + 21y \equiv \pm 15y \pmod{43}.$$
This gives us two diferent families of solutions:
$$x \equiv 37y \pmod{43} \qquad \text{and} \qquad x \equiv 7y\pmod{43}.$$
