Is $f(x,y)=\left(\frac{x^a(1-y)}{(1-x)}-\frac{y^{a}(1-x)}{(1-y)}\right)\frac{1}{x-y}$ convex? Is it possible to prove that,
$$f(x,y)=\left(\frac{x^a(1-y)}{(1-x)}-\frac{y^{a}(1-x)}{(1-y)}\right)\frac{1}{x-y}$$
is convex in the following range: $0<y<x<1$, where $a\ge2$ is an integer parameter.
Numerical testing shows that it is definitely convex, but can someone help to prove it?
If it is yet not convex I would be very thankful for a counter example. 
The Hessian is:
$\frac{\partial^{2}f}{\partial x \partial y}=\frac{xy^{a}(a(x-y)+2y)+x^{a}y(ax-2x-ay)}{xy(x-y)^3}$ 
$\frac{\partial^{2}f}{\partial x \partial x}=\frac{(y-1) x^{a+2} \left(a^2 (y (y+4)+1)-3 a (y (y+4)+1)+2 \left(y^2+y+1\right)\right)-2 (a-2) a y \left(y^2-1\right) x^{a+1}-2 (a-3) (a-1) \left(y^2-1\right) x^{a+3}+(a-3) (a-2) (y-1) x^{a+4}+(a-1) a (y-1) y^2 x^a-2 x^5 y^a+6 x^4 y^a-6 x^3 y^a+2 x^2 y^a}{(x-1)^3 x^2 (x-y)^3}$
according to Mathematica.
Below you can find some developments but the problem is not solved yet.

This is a partial answer.
Let us write
$$f(x,y) \:=\: \frac{x^a(1-y)^2-y^a(1-x)^2}{(1-x)(1-y)(x-y)}\tag{1}$$
and exploit that convex- or concavity is invariant under affine transformations.
Setting $\,u=1-y\,$ and $\,v=1-x\,$ leads to
$$\varphi(u,v,a) \:=\: \frac{u^2(1-v)^a-v^2(1-u)^a}{uv(u-v)}\tag{2}$$
with $\,0<v<u<1$.
As a side note: If $\,u=v>0\,$ then both the denominator and the numerator
vanish, hence these are removable discontinuities of this rational function. Furthermore, there's the symmetry $\,\varphi(v,u,a)=\varphi(u,v,a)\,$.
Expanding the $(1-\,?\,)^a$ terms in $(2)$, combining equal powers of $v$ and $u$, and simplifying then yields
$$\varphi(u,v,a) \:=\: \frac 1u+\frac 1v -a +\sum_{k=2}^{a-1}(-1)^k\binom a{k+1}
\underbrace{\frac{vu^k-uv^k}{u-v}}_{=\,\sum_{j=1}^{k-1}u^{k-j}v^j}\:.$$
Thus,
$$\varphi(u,v,a=2) \:=\:\frac 1u+\frac 1v -2$$
is convex because each summand is itself convex, and
$$\varphi(u,v,a=3) \:=\:\frac 1u+\frac 1v -3 +uv$$
is convex since its Hessian is
$$\begin{pmatrix}\frac 2{u^3} & 1\\ 1 & \frac 2{v^3}
\end{pmatrix} > 0$$
positive-definite.
Summary:
In the cases $a=2,3$ one gets that $f$ is a convex function.
The ansatz presented may pave the way for a complete answer
(which I currently do not see).
