Marden’s Theorem Given a triangle and three vertices in x, y format, is there a systematic way to use Marden’s Theorem to get the vertices of the foci of the inscribed ellipse?  It seems to involve the derivative but I cannot find any hardnumber examples online that clearly show how it works. 
 A: Just to be clear, there is no "the" inscribed ellipse. Marden's theorem tells you the foci of one particular inscribed ellipse (the one tangent to the midpoint of each side).
I think looking an example is probably clearer than trying to work it out in full generality. Say your triangle has vertices $(0,0)$, $(1,0)$, and $(0,1)$. Then the polynomial $p(z)=z(z-1)(z-i)$ has roots corresponding to the vertices of the triangle, so Marden's theorem tells you that the foci you're looking for are the roots of $p'(z)$.
Now,
$$p(z) = z^3-(1+i)z^2+iz \, ,$$
so
$$p'(z) = 3z^2-(2+2i)z+i \, .$$
Using the quadratic formula, this has roots
$$\frac{2+2i \pm \sqrt{(2+2i)^2-12i}}{6}=\frac{1+i \pm \sqrt{-i}}{3}=\frac{1 \pm \sqrt{2}/2}{3} + \frac{1 \mp \sqrt{2}/2}{3}i \, ,$$
so your foci are at $\left(\frac{1+\sqrt{2}/2}{3}, \frac{1-\sqrt{2}/2}{3}\right)$ and $\left(\frac{1-\sqrt{2}/2}{3}, \frac{1+\sqrt{2}/2}{3}\right)$.
Something similar should work in general -- you'll always get some cubic, differentiate it to get a quadratic, and be able to apply the quadratic formula.
If you want to find the foci of some other inscribed ellipse, the wikipedia article on Marden's theorem mentions a generalization (by Linfield) that you might be able to use, as long as you know something about the locations of its points of tangency.
