The general equation of an ellipse is x^2/a^2+y^2/b^2 where the origin or centre O is at (0,0); also the eccentricity is given by e, where b^2=a^2(1-e^2) and a and b are the x and y radii of the ellipse respectively (semi-major and semi-minor axes). The foci are at F(f,0) and F'(-f,0).
For any point P(x,y) on the ellipse PF+PF'=k a constant. That is, sqrt((x-f)^2+y^2)+sqrt((x+f)^2+y^2)=k. In particular, at the point Q(a,0) the extent of the x radius, so that P is at Q, sqrt((a-f)^2)+sqrt((a+f)^2)=k; a-f+a+f=k=2a. The shortest distance PF is when P is at Q, so that PF'=2PF (required by the question) and a+f=2(a-f); 3f=a; f=a/3.
At R(0,b) we have RF+RF'=k=2a; that is, 2sqrt(b^2+f^2)=2a, so b^2+f^2=a^2. The hypotenuses of the triangles ROF and ROF' have the same length as the radius a.
But f=a/3 and b^2=a^2(1-e^2), so a^2(1-e^2)+a^2/9=a^2 and 1-e^2+1/9=1, making e=1/3. The eccentricity is 1/3.