Solve $x,y,z$ subject to $$x^2+y^2 - xy = 3$$ $$(x-z)^2+(y-z)^2 - (y-z)(x-z) = 4$$ $$(x-z)^2+y^2 - y(x-z) = 1$$ $$x,y,z \in R^{+}$$
My attempts:
$(x-z)^2+(y-z)^2 - (y-z)(x-z) - ((x-z)^2+y^2 - y(x-z)) = xz - 2yz = (x-2y)z = 3$ $x^2+y^2 - xy - ((x-z)^2+y^2 - y(x-z)) = 2xz - yz - z^2 = (2x-y-z)z = 2$
divide above, we have $\frac{2x-y-z}{x-2y} = \frac{2}{3}$, or $3z = 4x+y$
Then the rest is easy. But I start to get interested in if there are some geometric solution. Like the comment says, they are the elliptical cylinders in the positive octant
We can take a quadrilateral $ABCD$, for which:
$AC\cap BD=\{E\},$ $AC=BD=z$,$ED=x$, $EC=y$, $\measuredangle AEB=60^{\circ},$$AB=2$, $BC=1$ and $CD=\sqrt3$.
Thus, we obtain your conditions: $$x^2+y^2-xy=ED^2+EC^2-2ED\cdot EC\cos60^{\circ}=CD^2=3,$$ $$(x-z)^2+(y-z)^2-(y-z)(x-z)=(z-x)^2+(z-y)^2-(z-x)(z-y)=$$ $$=EB^2+EA^2-2EB\cdot EA\cos60^{\circ}=AB^2=4$$ and $$(x-z)^2+y^2-y(x-z)=(z-x)^2+y^2+y(z-x)=$$ $$=EB^2+EC^2-2EB\cdot EC\cos120^{\circ}=1.$$ Now, let $BFCD$ be a parallelogram.
Thus, $$AC=z=BD=FC$$ and $$\measuredangle ACF=\measuredangle DEC=60^{\circ},$$ which gives that $\Delta AFC$ is equilateral triangle, $BA=2$, $BF=CD=\sqrt3$ and $BC=1.$
Now, let $R$ be a rotation around $C$ by $-60^{\circ}$ and $R\left(\left\{B\right\}\right)=\left\{G\right\}$ thus, since $\Delta BGC$ is an equilateral triangle, we obtain: $BG=1$, $FG=R(AB)=2$ and $$FG^2=2^2=(\sqrt3)^2+1^2=BF^2+BC^2,$$ which says $$\measuredangle FBG=90^{\circ},$$ $$\measuredangle FBC=\measuredangle FBC+\measuredangle GBC=90^{\circ}+60^{\circ}=150^{\circ},$$ which gives $$z^2=AC^2=FC^2=FB^2+BC^2-2FB\cdot BC\cos150^{\circ}=3+1+3=7,$$ which gives $$z=\sqrt7$$ and from here we get the answer: $$(x,y,z)=\left(\frac{5}{\sqrt7},\frac{1}{\sqrt7},\sqrt7\right).$$