Your work goes a long way towards the answer. I will assume that you are looking for solutions in real numbers $x$ and $y$. You want to solve $x^2-xy+y^2=0$. Note that $x^2-xy+y^2=\left(x-\frac{y}{2}\right)^2 +\frac{3}{4}y^2.\qquad(\ast)$ The above result is easy to verify by expanding the right-hand side. But it was not obtained by magic: It is a standard application of the powerful idea usually called Completing the Square. You have undoubtedly met this idea earlier in other contexts.
On the right-hand side of $(\ast)$ we have a square, namely $\left(x-\frac{y}{2}\right)^2$, plus $3/4$ of $y^2$. The square of the real number $y$ is always $\ge 0$. So the only way we can satisfy the equation $x^2-xy+y^2=0$ is by taking $y=0$ and $x-y/2=0$, meaning that $x=0$. These values are, as you pointed out, forbidden, so the original equation has no real solutions.
Remark: Through unhappy experiences, I have somewhat of an aversion to fractions, so would prefer to say that the equation $x^2-xy+y^2=0$ is equivalent to $4x^2-4xy+4y^2=0$. But $4x^2-4xy+4y^2=(2x-y)^2+3y^2.$ Or else, if we feel bad about breaking symmetry, we can avoid completing the square, and instead note that $2x^2-2xy+2y^2=(x-y)^2+x^2+y^2.$ Again, we have a sum of squares on the right, and this can be $0$ only if $x$, $y$ (and therefore $x-y$) are all $0$.
Much more mechanically, we can use the Quadratic Formula. For any fixed $y$, the solutions of $x^2-xy+y^2=0$ are $x=\frac{y \pm\sqrt{-3y^2}}{2}.$ If $y\ne 0$, the number under the square root sign is negative, and therefore $\sqrt{-3y^2}$ is not a real number, so $x$ is not a real number.