Prove that $x^3+2y^3+4z^3\equiv6xyz \pmod{7} \Rightarrow x\equiv y\equiv z\equiv 0 \pmod{7}$
Someone played a dirty trick on you.
The form $x^3-6xyz + 2y^3+4z^3$ is the “norm form” for the extension $\Bbb F_{7^3}\supset\Bbb F_7$. That is, if you take a generator $\zeta=\sqrt[3]2$ of the big field, which is all right, since $X^3-2$ is irreducible over $\Bbb F_7$, then the Norm of $x+y\zeta+z\zeta^2$ is $w=x^3-6xyz+2y^3+4z^3$, when $x,y,z\in\Bbb F_7$. But the Norm of a nonzero element of the big field is necessarily nonzero in the base field. And that does it.
How did I know this? Many hand computations of the Norm, over the years. Pencil and paper, friends, pencil and paper.
A more intrinsic answer:
The polynomial $x^3 - 2$ is irreducible over $\mathbb{F}_7$, hence induces a cubic extension $K = \mathbb{F}_7[\sqrt[3]{2}]$ of $\mathbb{F}_7$, with a basis $1, \sqrt[3]{2}, \sqrt[3]{4}$.
Now an element $\alpha = x + y\sqrt[3]{2} + z\sqrt[3]{4}$ of $K$ has norm $x^3 + 2y^3 + 4z^3 - 6xyz$. Hence this expression is zero if and only if the element $\alpha$ itself is zero, which means $x = y = z = 0$.
I warmly endorse the answers by WhatsUp and Lubin. I get the vibe that the machinery used in those is inaccessible to some readers, so here's an answer trying to meet in the middle.
Everything revolves around the following observation that can be verified by brute force checking (7 cases, all easy):
The congruence $u^3\equiv2\pmod 7$ has no solutions.
As a corollary to that we make the following observation.
Lemma 1. If $(x,y,z)\in\Bbb{Z}^3$ is a solution of the congruence $$ x^3+2y^3+4z^3-6xyz\equiv0\pmod7\qquad(*) $$ such that one of the variables is divisible by seven, so are the other two.
Proof. If $z\equiv0$ we are left with the congruence $$ x^3+2y^3\equiv0\Leftrightarrow x^3\equiv2(-y)^3.\qquad(**) $$ If $-y\not\equiv0$ it has a modular inverse $w$, i.e. an integer $w$ such that $w(-y)\equiv1$. Multiplying $(**)$ by $w^3$ then gives $(xw)^3\equiv2$ in violation of the Lemma. So $-y\equiv0$ implying $x^3\equiv0$, $x\equiv0$, and we are done with this case.
If $y\equiv 0$, then similarly $x^3\equiv4(-z)^3$. Multiplying this by $2$ gives the congruence $2x^3\equiv 8(-z)^3\equiv (-z)^3$. We get a similar contradiction by using a modular inverse of $x$ (that exists unless $x\equiv0$).
Leaving the case $x\equiv0$ as an exercise to the reader. QED.
Recall that we are working under the contrapositive assumption that $(x,y,z)$ is a solution of $(*)$ such that not all of them are zero. In light of Lemma 1. we can now assume that none of them are divisible by seven.
The next thing we can do is to use homogeneity. Because we assume that $z$ has a modular inverse $w$, we can multiply $(*)$ by $w^3$ and arrive at the congruence $$ (xw)^3+2(yw)^3+4-6(xw)(yw)\equiv0. $$ In terms of the new variables $u=xw$ and $v=yw$ this reads $$ u^3+2v^3+4-6uv\equiv0.\qquad(***) $$ In light of Lemma 1 we also know that neither $u$ nor $v$ is divisible by seven.
At this point we are left with the task of having to check $6^2=36$ pairs $(u,v)$, both ranging over $\{1,2,3,4,5,6\}$, instead of the initial $7^3-1=342$ triples.
I will testify that this is already doable in a timed setting. I recall having settled a similar three variable congruence modulo $7$ as an integral part of an IMO problem. I definitely had Lemma 1, but I don't remember, whether I figured out the use of homogeneity (effectively allowing us to assume $z=1$). I was young and determined at the time, and an IMO medal was on the line, so having to manually check 216 cases would not have stopped me either :-)
A further reduction involves the use of third roots of unity modulo seven (this actually already played a role in Lemma 1). Those are $2$ and $4$, as we have $$2^3\equiv 4^3\equiv2\cdot4\equiv1.$$ These congruences imply that $(u,v)$ is a solution of $(***)$ if and only if $(2u,4v)$ or $(4u,2v)$ are. It follows that it suffices to check all the pairs $(u,v)$ with $u\in\{1,2,3,4,5,6\}$, $v\in\{1,3\}$. This is because when $v$ ranges over $\{1,3\}$, between themselves $v$, $2v$ and $4v$ range over all of $\{1,2,3,4,5,6\}$.
We are down to 12 pencil & paper checks.