Answer
See below.
Work Step by Step
$x^2-xy+y^2=3\\
2x-(xy'+y)+2yy'=0\\
2x-xy'-y+2yy'=0\\
y'(2y-x)=y-2x\\
y'=\frac{y-2x}{2y-x}$
The ellipse crosses the $x$-axis when $y=0$
$x^2-xy+y^2=3\\
x^2=3\\
x=\pm\sqrt3$
Then we plug in $\pm\sqrt3$ into $x^2-xy+y^2=3$ and we get $(\sqrt3,0)$; $(-\sqrt3,0)$.
Now we need to check if the tangents at those points are parallel.
$(\sqrt3,0)$
$y'=\frac{y-2x}{2y-x}\\
y'=\frac{(0)-2(\sqrt3)}{2(0)-(\sqrt3)}\\
y'=2$
$(-\sqrt3,0)$
$y'=\frac{y-2x}{2y-x}\\
y'=\frac{(0)-2(-\sqrt3)}{2(0)-(-\sqrt3)}\\
y'=2$
Therefore, the tangent lines at $(\sqrt3,0)$ and $(-\sqrt3,0)$ are parallel as they both have the same gradient.
