The foci and directrices are symmetic with respect to the minor axis, so once you have the center, you can find the other focus and directrix by reflection in the line through this point parallel to the known directrix. The following solution doesn’t require first finding the equation of the ellipse.
Let $D$ be the intersection of the directrix and its perpendicular through the focus $F$ (i.e., with the major axis of the ellipse). One vertex $A$ of the ellipse is between these two points. For this vertex, $|A-F|=e|D-A|$, so $A={1\over1+e}F+{e\over1+e}D$. For the other vertex $A'$, we have $|A'-F|=e|A'-D|$, from which $A'={1\over1-e}F-{e\over1-e}D$. The center of the ellipse is their midpoint $$C=\frac12(A+A')={1\over1-e^2}F-{e^2\over1-e^2}D.$$ In this problem, $F=(0,0)$ and $D=(-1,0)$, so the center of the ellipse is at $${1\over1-\left(\frac12\right)^2}(0,0)-{\left(\frac12\right)^2\over1-\left(\frac12\right)^2}(-1,0)=-\frac14\cdot\frac43\cdot(-1,0)=\left(\frac13,0\right).$$ From this, we easily find that the other focus is $\left(\frac23,0\right)$ with corresponding directrix $x=\frac53$.
If you have the equation of the conic in hand, you can find its center by differentiating and setting the partial derivatives to zero, i.e., $$\frac32x+\frac12=0\\ \frac32y=0$$ which has solution $x=-\frac13$, $y=0$. This point is between the given focus and directrix, so as I mention in my comment to your question, this means that the equation that you’ve come up with is incorrect. That aside, once you’ve found the correct center point $(c,0)$, the other focus is the reflection in this point, namely, $(2c,0)$, with corresponding directrix $x=2c+1$.
Since you know that the points are colinear and you know their distances from the midpoint, a simple way to find the vertices is to compute the vectors from the midpoint to the foci and scale them to have the right length. You’ve got the midpoint $(5,2)$, so the two vectors are $(2,6)-(5,2)=(-3,4)$ and its negative. The $c$ value is $5$, so these vectors have length $5$ (you can check that for yourself). You need vectors of length $9$: scale them appropriately and add them to the midpoint to get the two vertices.
Best Answer
I'm assuming there is some educational reason why you're not supposed to use the formula $d_1+d_2=2a$. That formula gives the correct result for good reasons.
The method in which you solve two simultaneous quadratic equations is legitimate; after eliminating $y$, we can show that $x(x-2)=0$. If we allow complex solutions, this leads to four solutions; but since the points on the ellipse have real coordinates, it is perfectly legitimate to discard the complex solutions. The fact that $x$ and $y$ must be real is simply a constraint on the solution space; you can even add it to your system of equations as follows: \begin{gather} \tfrac94(x-x_0)^2 + 3y^2 = 1,\\ x^2 + y^2 = 1,\\ \Im (x) = 0,\\ \Im (y) = 0. \end{gather} The last two equations simply say that $x$ and $y$ have zero imaginary parts.
Another technique is to convert the equation of the ellipse to polar coordinates. The ellipse equation you were given was in the general form $$ \frac{(x-x_0)^2}{a^2}+\frac{(y-y_0)^2}{b^2}=1 $$ for positive $a$ and $b$, where in this particular case $x_0=\frac13$, $y_0=0$, $a=\frac23$, and $b=\frac{1}{\sqrt3}$. Since $\frac23 > \frac{1}{\sqrt3}$, the length of the semi-major axis is $a$. In polar coordinates, the general equation for an ellipse with semi-major axis $a$, semi-minor axis $b$, one focus at $(0,0)$, and the other focus on the positive $x$-axis, like this ellipse, is $$ r = \frac{a(1 - e^2)}{1 - e\cos\theta} $$ where $e = \sqrt{1 - \left(\frac ba\right)^2}$. Plugging in $a=\frac23$ and $b=\frac{1}{\sqrt3}$, we find that $e = \sqrt{1 - \frac34} = \frac12$, so the equation of this particular ellipse comes out to $$ r = \frac{\frac23\left(1 - \left(\frac12\right)^2\right)} {1 - \frac12\cos\theta} = \frac{1}{2 - \cos\theta}. $$ But since $r$ in this equation is simply the distance from $(0,0)$ to a point on the ellipse at angle $\theta$ counterclockwise from the positive $x$-axis, and the desired point $P$ is at distance $\frac12$ from $(0,0)$, the polar coordinates of $P$ must satisfy $$ \frac12 = \frac{1}{2 - \cos\theta}, $$ from which we can deduce that $\cos\theta = 0$, and therefore $\sin\theta = \pm 1$. Converting the polar coordinates back into $x,y$ coordinates gives us \begin{align} x &= r \cos\theta = 0,\\ y &= r \sin\theta = \pm\frac12. \end{align} Therefore $P = \left(0,\frac12\right)$ or $P = \left(0,-\frac12\right)$.