If you really mean $d\ge x$, then $f(x,y)\le f(d,y)$ which is convex, or even $f(x,y)\le f(d,0)$.
If it was a typo and the true domain is $[d,+\infty)\times[0,+\infty)$, the convex function $g(x,y):=f(x,0)+f(d,y)-f(d,0)=\frac{x}{c}+\frac{d}{c+y}-\frac{c}{d}$ has $g(d,y)=f(d,y)$ for all $y\ge 0$ and $g_x(x,y)\ge f_x(x,y)$ for all $x\ge d$ and $y\ge0$ , so $g\ge f$ on $[d,+\infty)\times[0,+\infty)$.