<s>[INST] Find $f(2)$ if $f$ satisfies $2f(x)-3f(\frac1x)=x^2$ The following expression is given, and we are asked to find $f(2)$. 
\begin{equation}
2f(x)-3f\left(\frac{1}{x}\right) =x^2
\end{equation}
Does a unique and well defined answer exist? Why? and what is it? 
We have that $f(1)=f(-1)=-1$. 
 [/INST] Hint: Note that
$$
\begin{bmatrix}
2&-3\\
-3&2
\end{bmatrix}
\begin{bmatrix}
f(x)\\
f(1/x)
\end{bmatrix}
=
\begin{bmatrix}
x^2\\
1/x^2
\end{bmatrix}
$$
and
$$
\det\begin{bmatrix}
2&-3\\
-3&2
\end{bmatrix}=-5\ne0
$$

Completed Answer
Now that over a year has passed, I think it is appropriate to finish the answer.
$$
\begin{align}
\begin{bmatrix}
f(x)\\
f(1/x)
\end{bmatrix}
&=
\begin{bmatrix}
2&-3\\
-3&2
\end{bmatrix}^{-1}
\begin{bmatrix}
x^2\\
1/x^2
\end{bmatrix}\\
&=
-\frac15\begin{bmatrix}
2&3\\
3&2
\end{bmatrix}
\begin{bmatrix}
x^2\\
1/x^2
\end{bmatrix}\\
&=-\frac15\begin{bmatrix}
2x^2+3/x^2\\
3x^2+2/x^2
\end{bmatrix}\\
\end{align}
$$
Therefore,
$$
f(x)=-\frac{2x^4+3}{5x^2}
$$
</s>