It's easy to show that if $x=y=z$ or two of them are equal and the other is zero, then the equality holds. What about the other direction?
EDIT: I think this is more like a combinatorial problem. To prove if the equality holds, then $x=y=z$ or two of them are equal and the other is zero. That is $$x^t(x-y)(x-z) + y^t(y-z)(y-x)+z^t(z-x)(z-y)=0 \implies (x=y=z) \text{ or } (x=y \text{ and } z=0)\\ \equiv \cdots \implies (x=y \text{ and } y=z \text{ and } x=z) \text{ or } (x=y \text{ and } z=0).$$
To prove the contrapositive, we need to prove $$\neg(x=y \text{ and } y=z \text{ and } x=z) \text{ and } \neg(x=y \text{ and } z=0) \implies x^t(x-y)(x-z) + y^t(y-z)(y-x)+z^t(z-x)(z-y)=0\\ \equiv (x\not =y \text{ or } y\not =z \text{ or } x\not =z) \text{ and } (x\not =y \text{ or } z\not =0) \implies x^t(x-y)(x-z) + y^t(y-z)(y-x)+z^t(z-x)(z-y) \not=0.$$ There are $(3+3+1)(2+1)=21$ possibilities of the hypothesis and we also need to consider the supposition, say, $x\geq y\geq z\geq 0$, of the proof. Hence, we need to find a way to combine some cases to get a concise proof. Am I correct?
You seem to want to show the implication that if $x^t(x-y)(x-z) + y^t(y-z)(y-x)+z^t(z-x)(z-y)=0$ for $x, y, z \geqslant 0$ and $t>0$ then at least two among $x, y, z$ are the same and the other zero, or all are equal.
First suppose all are distinct, i.e. WLOG $x>y> z\geqslant 0$ using symmetry. Then we may divide throughout by the positive number $(x-y)(y-z)(x-z)$ to get $$\frac{x^t}{y-z}+\frac{z^t}{x-y}=\frac{y^t}{x-z}$$ Note that all the three fractions are necessarily non-negative. Further, $x^t> y^t$ and $y-z < x-z$, so the first term on LHS is strictly higher than the RHS. Hence this equality is not possible, i.e. $x, y, z$ cannot be distinct.
Now we consider that perhaps two among the three are equal, i.e. $y=z$, say. In this case the equality becomes $x^t(x-y)^2=0$, which is satisfied iff $x=0$ or $x=y$.
Hence we have proved the implication required.