$1/i=i$. I must be wrong but why? [duplicate]

What you are doing is a version of $$ -1=i^2=\sqrt{-1}\sqrt{-1}=\sqrt{(-1)(-1)}=\sqrt1=1. $$ It simply shows that for non-positive numbers, it is not always true that $\sqrt{ab}=\sqrt{a}\sqrt{b}$.

$$\frac1{\sqrt{-1}}=\sqrt{-1}$$ is only true in the sense that $1$ over a square root of $-1$ is a square root of $-1$. However, there are two square roots of every non-zero complex number, so you have to make sure to pick the right one. For square roots of non-negative real numbers it works to consistently pick the non-negative square root, but no such rule exists for all complex numbers.

You assume $\sqrt{-1}=i$, but it's wrong to write such a thing, because the square root function is really defined as a function only for positive argument, and it has positive values. If you write everywhere $\pm \sqrt{a}$ instead of $\sqrt{a}$ (because it really means "a root of $x^2-a$" and it has two roots), then everything you wrote is "almost right".

Even for real numbers, there are two square roots. $(-2)^2 = 2^2 =4$, so both $2$ and $-2$ can be thought of as square roots of $4$. In the real numbers, we have an easy way to pick one of the two: just always pick the positive one. So we define $\sqrt{x}$ to be the positive square root of $x$. In the complex case, we do not have an order, and so no consistent way to pick one of the two square roots. So $\sqrt{z}$ is not really a function: it is a multivalued function. You confusion arises from thinking that $\sqrt{i}$ indicates only one number.

It is wrong because the calculation rule of square roots only works for real non negative roots.