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I would like to write following sentence in second order logic:
universum has even number of elements.
And my atempt is as follows:
$$\psi \wedge \exists_{f()} \forall_x \forall_y x\neq y \wedge f(x) = y \rightarrow f(y) = x $$ $\psi$ states that universum is finite - it is easy to express in second order logic.

Is this correct ?

Edition Below, @Noah answered and corrected slightly my solution. However, I would like to try use his hint for alternative solution:
$$\phi_A=\exists_{A_1}\exists_{A_2}\forall_{x\in A}\left[ (x\in A_1\rightarrow x\notin A_2 \wedge x\in A_2\rightarrow x\notin A_1\right) \wedge (\exists_{f:A_1\to A_2} \text{f is injective and surjective}]$$

$$\phi_A \wedge (\forall_{x\in A}\neg \phi_{A\setminus \{x\}})$$ expresses weakly even - I believe. However, I don't know how to express strong parity. Can you help me ?

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This is almost correct. You need, however, to rule out the possibility of $f(x)=x$; e.g. the following would work: $$\psi\wedge\exists F\forall x\forall y[(F(x)=y\implies x\not=y)\wedge(F(x)=y\implies F(y)=x)].$$

Another way to do this, which I think is kind of neat since it avoids invoking $\psi$, is the following. Say a set is weakly even if it is finite and of even size, or is infinite; basically, a set is even if it can be partitioned into pairs. Then a set $X$ is (genuinely) even iff $X$ is weakly even, and there is no element $x\in X$ such that $X\setminus \{x\}$ is also weakly even. But "weakly even" can be described similarly to what we've done above! So this yields an expression of "the universe is (finite and) even" in second-order logic, which doesn't involve $\psi$.

Interestingly, this second approach requires the axiom of choice! It is consistent with ZF (= set theory without choice) that there is an infinite set which can be partitioned into pairs, but which cannot be partitioned into pairs + one leftover element. So this is a sense in which the approach you've taken is better.

EDIT: Your expression does indeed describe weak evenness. And it's easy to tweak it to get strong evenness: just add $\psi$, as before.

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    I edited my post. – 2017-02-02
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    @HaskellFun I've edited mine. – 2017-02-02
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    To EDIT: Yes, I know that I can add prevention $\psi$ for forcing finite set, however, you told somehting that I can avoid using $\psi$ – 2017-02-02
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    @HaskellFun Well, you can do the same thing here: just add a sentence saying that $A$ minus one element *isn't* weakly even. *Again, this only works assuming Choice.* – 2017-02-02
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    So in $\phi$ I should write: $\exists_{A_1}\exists_{A_2}\forall_{x\in A}...$ Yeah ? – 2017-02-02
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    @HaskellFun Isn't that what you've already written? – 2017-02-02
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    Look more exatly on $\forall$ – 2017-02-02
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    @HaskellFun I don't know what "$A$" is. If it's the whole universe, then "$\forall_{x\in A}$" and "$\forall_x$" mean exactly the same thing. If it's not, what is it? – 2017-02-02
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    Ok, You are right. Look at my editions - in a moment I add them – 2017-02-02