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I am considering the following function
$$f(x_1,dots,x_n,y)=-alpha left(y-k_1right)^2-beta sum_{i=1}^nleft(k_2-x_iright)^2-gamma sum_{i=1}^nleft(y-x_iright)^2 - frac{delta}{y-d} sum_{i=1}^n (x_i-d), ,$$
where $(x_1,dots,x_n,y)in[d,1]^{n+1}$, $d>0$ and $alpha$, $beta$, $gamma$, $delta$, $k_1$, $k_2>0$. Moreover, $x_ileq y$ $forall$ $i=1,dots,$ $n$. I'm trying to calculate the maximum of this function on that domain.

Befor using "brute force" approach (i.e. by calculating derivatives, Hessian and so on), I wonder whether it's possible to obtain the absolute maximum in a more simple way. For example, I notice that $fleq0$...

Consider $y$ fixed as Paul Sinclair suggested. Your objective function is now separable: you can optimize each $x_i$ independently. It is also concave in $x_i$ (since the second derivative is negative), so it is maximized when the derivative is 0:
$$-2beta (k_2-x_i) - 2gamma(y-x_i) - frac{delta}{y-d} = 0$$
which can be written as
$$2(beta + gamma)x_i = 2beta k_2 + 2gamma y + frac{delta}{y-d}$$
so the solution is
$$x_i = frac{beta k_2 + gamma y}{beta + gamma} + frac{delta}{2(beta + gamma)(y-d)}$$
You can plug this in and maximize over just $y$. Since the last term in your objective is not squared, maximizing over $y$ is not simple: there are probably multiple local optima. You could perform grid search for $y$, or apply a gradient based optimization algorithm and try multiple starting points.

Are really the derivatives way the bruit force?

The derivates of
$$f(x_1,dots,x_n,y)=-alpha left(y-k_1right)^2-beta sum_{i=1}^nleft(k_2-x_iright)^2-gamma sum_{i=1}^nleft(y-x_iright)^2 - frac{delta}{y-d} sum_{i=1}^n (x_i-d), $$
are zeros in the stationary points,
$$begin{cases}
f''_{x_j}(x_1,dots,x_n,y)=2beta left(k_2-x_jright)+2gamma left(y-x_jright) - dfrac{delta}{y-d}, =0\
f'_y(x_1,dots,x_n,y)=-2alpha left(y-k_1right)-2gamma sumlimits_{i=1}^nleft(y-x_iright) - dfrac{delta}{y-d} sumlimits_{i=1}^n (x_i-d), =0,
end{cases}$$
or, for $ynot=d,$
$$begin{cases}
2beta (y-d)left(k_2-x_jright)+2gamma (y-d) left(y-x_jright) - delta, =0, quad j=1dots n\
-2alpha (y-d)^2left(y-k_1right)-gamma (y-d)^2sumlimits_{i=1}^nleft(y-x_iright) - delta sumlimits_{i=1}^n (x_i-d), =0,
end{cases}$$
so
$$begin{cases}
-2(beta+gamma) (y-d)x_j + 2(y-d)(beta k_2+gamma y) - delta, =0, quad j=1dots n\
-2(beta+gamma) (y-d)sumlimits_{i=1}^n x_i + 2n(y-d)(beta k_2+gamma y) - ndelta, =0\
(gamma-delta)(y-d)^2sumlimits_{i=1}^nx_i + (2alpha k_1 - gamma y)n(y-d)^2 +ndelta d, =0.
end{cases}$$
Summation of the second and third equations factors $2(gamma-delta)(y-d)$ and $(beta+gamma)$ gives
$$begin{cases}
-2(beta+gamma) (y-d)x_j + 2(y-d)(beta k_2+gamma y) - delta, =0, quad j=1dots n\
2(gamma-delta)(2n(y-d)(beta k_2+gamma y) - ndelta) + ((2alpha k_1 - gamma y)n(y-d)^2 +ndelta d)(beta-gamma), =0,
end{cases}$$
and this leads to the cubic equation for $y$ and explicit expressions for $x_j.$