to simplify the notations i’m using input = x, output = y, target = t, weight = w

If I understand correctly, the following is not generally true,

Off course, this thig has no sens when at least one element of x is close to 0. but what you can do is to change your target, since you want y(w,x) = xt + c, by playing with w and c. So you can detach x and put it into your target:
min criterion[y(w,x)-c, tx] by playing with w and c

Sorry for not making it clear.
Actually, If we backward (output-c)/input, we get
[…]
needed by the optimizer.

The thing is, what do you want for your equilibria ? if you want to play with w in order to get dy/dx = t, then both methods converge to this equilibria. With my technic, you will play with w and c in orger to get y = t*x + c for all (x,t)
You can verify that the two equilibria are equivalent.
And off course, d( dy/dx )/dw is different than d( y )/dw. Thins is not a probleme, we are no longer trying to minimze the same functions. The only thing that count is what we obtain after convergency.

I don’t think there is work-around for the second derivative. So I’ll keep an eye on the topic you mentioned.

Don’t give up ! You don’t need second order derivative !