}

Equation (2) is equivalent to \(\mathrm{to} \frac{1}{x} \cdot f\left(\frac{1}{x}\right)=x^{3} \cdot g(x)\).

If one performs the substitution \(x \leftarrow \frac{1}{x}\) in the first equation, one obtains \(\frac{1}{x} \cdot f\left(\frac{1}{x}\right)=1+x \cdot g(x)\), thus together

\[
1+\left(x-x^{3}\right) \cdot g(x)=0 \quad \text { or } \quad g(x)=\frac{1}{x^{3}-x}
\]

and

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

thus \(x \cdot f(x)=\frac{1}{1-x^{2}}\) and therefore

\[
f(x)=\frac{1}{x-x^{3}}=-g(x)
\]
\section*{Problem 4 - 131244}

Determine all pairs \((f, g)\) of functions that are defined for all real numbers \(x\) except \(-1, 0,\) and \(1\), and satisfy the following equations for all such \(x\):

\[
\begin{aligned}
x \cdot f(x) - \frac{1}{x} \cdot g\left(\frac{1}{x}\right) & = 1 \\
\frac{1}{x^{2}} \cdot f(x) & = x^{2} \cdot g(x)
\end{aligned}
\]