Find all functions $f: \mathbb{R}\setminus\{0\} \to \mathbb{R}$ satisfying $3f(x) - f\left(\frac{1}{x}\right) = x^2$.
What I did was first plug in $x = 1$ to get $2f(1) = 1 \implies f(1) = \frac{1}{2}$. Then seeing as how this looks symmetric I did the substitution $x \mapsto \frac{1}{x}$ to get $3f(\frac{1}{x})-f(x) = \dfrac{1}{x^2}$ then we can solve for $f(x)$ to get $8f(x) = 3x^2+\dfrac{1}{x^2} \implies f(x) = \frac{1}{8}(3x^2+\frac{1}{x^2})$.
This is indeed correct. The key thing to notice that $f(x)=\frac{1}{x}$ is an involution (i.e. $f(f(x))=x$) and whenever you notice that a certain function has this $f^n(x)=x$ property, its very useful to set $x$ to $f(x)$ and see what happens. Often, once you do this, you will end up with a system of solvable simultaneous equations involving variables like $f(x),f(\frac{1}{x}),f(x-1),\dots$ allowing you to solve directly for $f(x)$ as you would a normal system of equations.
