Q:

Joint Convexity Proof

Let $x$ be $n \times 1$ vector and $Y$ be $n \times n$ matrix. 
Prove that $f(x,Y) = x'Y^{-1}x$ is jointly convex in $x$ and $Y$ when $Y \succ 0$.

A:

Are you leaving out some assumptions?  What if $n=1$ and $Y$ is negative?
EDIT: With the new assumptions, we may consider a line segment through $(x,Y)$ space: $x = x_0 + t x_1$, $Y = Y_0 + t Y_1$, where $Y_0$ is positive definite and $Y_1$ is symmetric.  It is enough to show that 
$\left.\dfrac{d^2}{dt^2} (x' Y^{-1} x)\right|_{t=0} \ge 0$.  Now if $Z = Y_0^{-1/2} Y_1 Y_0^{-1/2}$, $w_0 = Y_0^{-1/2} x_0$ and $w_1 = Y_0^{-1/2} x_1$, 
$$Y^{-1} = (Y_0^{1/2} (I + t Z) Y_0^{1/2})^{-1} = Y_0^{-1/2} (I - t Z + t^2 Z^2 + \ldots) Y_0^{-1/2}$$
so
$$ x' Y^{-1} x = (w_0 + t w_1)' (I - t Z + t^2 Z^2 + \ldots) (w_0 + t w_1)$$
and $$\eqalign{\left.\dfrac{1}{2} \dfrac{d^2}{dt^2} (x' Y^{-1} x)\right|_{t=0} &=
w_0' Z^2 w_0 - w_1' Z w_0 - w_0' Z w_1 + w_1' Z^2 w_1 \cr
&= (Z w_0 - w_1)'(Z w_0 - w_1) \ge 0}$$
Thus it is indeed jointly convex.

