If the conditions of the problem hold for an $x \in \mathbb{R}$, then they also hold for $-x$. By substituting into

$$
2 f\left(\frac{x^{2}+x-1}{x^{2}-x-1}\right)-3 f\left(\frac{x^{2}-x-1}{x^{2}+x-1}\right)=5\left(x-\frac{1}{x}\right)
$$

we obtain

$$
2 f\left(\frac{x^{2}-x-1}{x^{2}+x-1}\right)-3 f\left(\frac{x^{2}+x-1}{x^{2}-x-1}\right)=-5\left(x-\frac{1}{x}\right)
$$

thus forming a linear system of equations in $f\left(\frac{x^{2}+x-1}{x^{2}-x-1}\right)$ and $f\left(\frac{x^{2}-x-1}{x^{2}+x-1}\right)$, from which it follows that

$$
f\left(\frac{x^{2}+x-1}{x^{2}-x-1}\right)=x-\frac{1}{x}
$$

Now, (3) can be written using the mappings

$$
\tilde{f}: \mathbb{R} \backslash\{1\} \rightarrow \mathbb{R} \quad \text { with } \quad x \mapsto \frac{x+1}{x-1}
$$

and

$$
g: \mathbb{R} \backslash\{0\} \rightarrow \mathbb{R} \quad \text { with } \quad x \mapsto x-\frac{1}{x}
$$

in the form

$$
f(\tilde{f}(g(x)))=g(x) \quad(x \in \mathbb{R} \backslash\{0\})
$$

which is easily verified. Since $g(x)$ runs through all real numbers for $x \neq 0$, it must follow that for all real numbers $x \neq 1$, $f(\tilde{f}(x))=x$ must hold. On the other hand, $\tilde{f}(\tilde{f}(x))=x$ also holds for all $x \neq 1$, as can be easily verified, which implies that $f(\tilde{f}(x))=\tilde{f}(\tilde{f}(x))$. Consequently,

$$
f(x)=\tilde{f}(x)=\frac{x+1}{x-1} \quad(x \in \mathbb{R} \backslash\{1\})
$$

Finally, by substituting into (1), one can verify that this is indeed a solution to our functional equation.
## Task 3A - 321233A

Determine all functions $f$ that are defined for all real numbers $x$ with $x \neq 0$ and $x \neq 1$, and for all real numbers $x$ with $x \neq 0$, $x^{2}-x-1 \neq 0$, and $x^{2}+x-1 \neq 0$, and satisfy the following equation (1):

$$
2 \cdot f\left(\frac{x^{2}+x-1}{x^{2}-x-1}\right)-3 \cdot f\left(\frac{x^{2}-x-1}{x^{2}+x-1}\right)=5 \cdot\left(x-\frac{1}{x}\right)
$$