My question is the function $(x,y)\mapsto f(x,y)$ convex where
$$f(x,y)=x\log\left(1+\frac{1}{1+y}\right).$$
I have found the partial derivatives as follow:
$$\dfrac{\partial^2 f(x,y)}{\partial x^2}=0.$$ $$\dfrac{\partial^2 f(x,y)}{\partial y^2}=\dfrac{x(3+2y)}{(1+y)^2(2+y)^2}.$$ $$\dfrac{\partial^2 f(x,y)}{\partial x\partial y}=-\dfrac{1}{(1+y)(2+y)}.$$
But I cannot continue.
The function $y \mapsto \ln(1+ {1 \over 1+y} )$ is not convex for $y<2$, hence $f$ is not convex on $\{1\} \times (-\infty, -2)$.