I should probably use the fact that $r/d = e$, where $r$ is the distance from the focus to any point $M(x,y)$ of an ellipse. $d$ the distance from $M(x,y)$ to the directrix, and $e$ is the eccentricity.
If you showed your attempt to begin with, we might be able to be a bit more helpful; alas, since you haven't shown it, you'll have to content yourself with the following sketch of a solution.
You know the distance from an arbitrary point $(x,y)$ to the focus $(3,0)$:
$f=\sqrt{(x-3)^2+(y-0)^2}$
and you can use the formula for point-line distance (formula 11 here) to get the distance from $(x,y)$ to the line $x+y-1=0$:
$d=\frac{x+y-1}{\sqrt{1^2+1^2}}$
from which you use the definition for eccentricity, $\varepsilon=\dfrac{f}{d}$, where $\varepsilon=\dfrac12$.
At once you should obtain an equation with a square root. You can try squaring both sides of the equation and then rearrange things to obtain a two-variable quadratic as usual, but you'll have to justify why the squaring is legal. You should end up with
$7x^2-2xy+7y^2-46x+2y+71=0$