Do I need to verify solutions to functional equations?

I am studying the (basics of) solving functional equations. My teacher stipulates that we check any solutions obtained by substitution. Similar guidelines are given in this IMO training material.

For a typical example, consider $f:\mathbb{R} \setminus \{0\} \mapsto \mathbb{R}$

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

The substitutions $x \mapsto y$ and $x \mapsto \frac{1}{y}$ yield (after some algebra) the solution $f(y) = \frac{2y}{3} - \frac{1}{3y}$. Verifying, we see this is indeed a solution.

My question: In algebra we have some well-known possibilities how extraneous solutions can appear. Are there additional ways of obtaining extraneous solutions in functional equations?

Answer

Here is an example of where we must check our solution into the equation to find the correct answer.

Find all functions $f:\mathbb{Z}\to \mathbb{Z}$ such that $$f(x+f(y)+xf(y))=x+xy+f(y)$$

Solution:
Notice that the LHS is symmetric - if we let $f(a)=x$, then swapping $a$ and $y$ keeps it the same. But the RHS isn't symmetric, hence substituting $(x,y)=(f(a),y)$ and then swapping $a$ and $y$ equates both RHS's - hence $$f(a)y=f(y)a$$ for all integers $a,y$. Then substituting $a=1$ shows $f(y)=yf(1)$, so $f(x)=cx$ for some constant $c=f(1)$. But substituting this into the original question shows that $c$ must be $1$. Hence $f(x)=x$ is our only function.