Doubt: $f\bigl(f(x)f(y)\bigr) + f(x+y) = f(xy) $ over $\mathbb R$ 
Let $\mathbb{R}$ be the set of real numbers. Determine all functions $f: \mathbb{R} \to \mathbb{R}$ such that for any real numbers $x$ and $y$, $$f\bigl(f(x)f(y)\bigr) + f(x+y) = f(xy) \text.$$

This is from Evan Chen's IMO 2017 Solution Notes:

The only solutions are $f(x) = 0$, $f(x) = x − 1$ and $f(x) = 1 − x$, which clearly work. Note that If $f$ is a solution, so is $−f$.

I understood how $f(x) = 0$, $f(x) = x − 1$ and $f(x) = 1 − x$ work.
When $f(x)=0$, we get
$$f\bigl(f(x)f(y)\bigr) + f(x+y) = f(0)+0=0=f(xy) \text.$$
When $f(x)=x-1$, we get
$$f\bigl(f(x)f(y)\bigr) + f(x+y) = f\bigl((x-1)(y-1)\bigr)+ x+y-1 \\
= xy-x-y+x+y-1=xy-1= f(xy) \text.$$
When $f(x)=1-x$, we get
$$f\bigl(f(x)f(y)\bigr) + f(x+y)= f\bigl((1-x)(1-y)\bigr) + 1 -x-y \\
= -xy +x+y +1 -x-y= 1-xy  f(xy) \text.$$

Can someone explain this line "If $f$ is a solution, so is $−f$".
Thanks in advance!

Let $g=-f$ with $f$ a solution to the equation. Then $$g(xy)=-f(xy)=-f(f(x)f(y))-f(x+y)\\ =g(-g(x)(-g(y)))+g(x+y) = g(g(x)g(y)) +g(x+y)$$
and so $g$ also satisfied the functional equation.
