Function $f(x;f(y;z))=k\cdot f(f(x;y);z)$ Is there function $f: \mathbb R^2 \rightarrow\mathbb R $ such that:

*

*$\forall a \in \mathbb R$ $\exists x, y \in \mathbb R$ such that $f(x;y)=a$


*$\forall  x, y \in \mathbb R$
$$f(x;y)=-f(y;x)$$


*$\forall  x, y, z \in \mathbb R$
$$f(x;f(y;z))=2021f(f(x;y);z)?$$
###My work so far:
I have proved
I. $\forall a \in \mathbb R$
$$f(a;a)=0$$
II. $\forall a \in \mathbb R$
$$f(a;0)=f(0;a)=0$$
III. I have hypothesis that $f(a;1)=a$ for any $a$ in $\mathbb R$ but I think there are not exist such function
 A: This is my try: 
Assume that there exists $f: \mathbb R^2 \rightarrow\mathbb R $ a function such that:
(1) $\forall a \in \mathbb R$ $\exists x, y \in \mathbb R$ such that $f(x;y)=a$
(2) $\forall  x, y \in \mathbb R$
$$f(x;y)=-f(y;x)$$
(3) $\forall  x, y, z \in \mathbb R$
$$f(x;f(y;z))=2021f(f(x;y);z)$$
Let $(a, b) \in \mathbb{R}^2$, then thanks to (1) there exists  $(x, y) \in \mathbb{R}^2$ such that $b=f(x, y) $. Thus
$$f(a, b)=f(a, f(x, y))...(4)$$
Now, we have
$$\begin{array}{ccc} 
f(a, f(x, y))&=&-f(f(x, y), a) \hspace{0.5cm}\text{(from(2))} \\
&=&-\dfrac{1}{2021} f(x, f(y, a)) \hspace{0.5cm}\text{(from(3))} \\
&=&\dfrac{1}{2021} f(f(y, a),x) \hspace{0.5cm}\text{(from(2))}    \\
&=&\dfrac{1}{2021^2} f(y, f(a,x))    \hspace{0.5cm}\text{(from(3))}   \\
&=&-\dfrac{1}{2021^2} f(f(a,x),y)    \hspace{0.5cm}\text{(from(2))}   \\
&=&-\dfrac{1}{2021^3} f(a,f(x, y))    \hspace{0.5cm}\text{(from(3))}   \\
\end{array} $$
Then
$$ \left(1+\dfrac{1}{2021^3} \right) f(a,f(x, y))=0 $$
which gives
$$ f(a,f(x, y))=0 $$
Now, $(4)$ allows to say that
$$ \forall (a, b) \in \mathbb{R}^2; f(a, b)=0$$
but this function doesn't satisfy (1), hence we have no solutions.
