Let $\mathbb{R}$ be the set of real numbers. Determine all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that for all real numbers $x, y$,

$$
f(f(x) f(y))+f(x+y)=f(x y)
$$
Matolcsi Dávid's solution. If $f(0)=0$, then for $y=0$ we get:

$$
f(f(x) f(0))+f(x+0)=f(0 \cdot x)
$$

In this case, $f(x)=0$ for all $x$. This is indeed a solution to the functional equation.

Now let's consider the case when $f(0) \neq 0$. If $x=0$ and $y=0$, then

$$
f\left(f(0)^{2}\right)+f(0)=f(0), \quad f\left(f(0)^{2}\right)=0
$$

Suppose $f(c)=0$ and $c \neq 1$. If $y=1+\frac{1}{x-1}$, then $(x-1)(y-1)=1$, i.e., $x+y=xy$, so $f(x+y)=f(xy)$, and thus $f(f(x) f(y))=0$.

Let $x=c$ and $y=1+\frac{1}{c-1}$. Then $f\left(f(c) f\left(1+\frac{1}{c-1}\right)\right)=0$. Since $f(c)=0$ and $f(0)=0$, this contradicts the initial condition.

We have found that if $f(c)=0$, then $c=1$, so $f(0)^{2}=1$, hence $f(1)=f\left(f(0)^{2}\right)=0$, and $f(0)=1$ or $f(0)=-1$.

It is clear that $f(x)$ is a solution to the functional equation if and only if $-f(x)$ is also a solution (the sign of both sides changes). Therefore, without loss of generality, we can assume that $f(0)=-1$.

Now substitute $y=1$ into the equation:

$$
\begin{aligned}
f(f(x) f(1))+f(x+1) & =f(1 \cdot x) \\
f(0)+f(x+1) & =f(x) \\
f(x+1) & =f(x)+1
\end{aligned}
$$

From this, it follows that for integers $n$, $f(x+n)=f(x)+n$.

We will show that $f(x)$ is injective. Suppose it is not, i.e., there exist $A \neq B$ such that $f(A)=f(B)$. Let $n$ be an integer greater than $A$, and let $A-n=a$ and $B-n=b$, where we know that $a$ is negative.

If $f(A)=f(B)$, then $f(a)=f(b)$. The discriminant of the quadratic equation $x^{2}-b x+a-1=0$ is $b^{2}-4(a-1)$, which is positive (since $a-1$ is negative), so the equation has two roots, $r$ and $s$. From Vieta's formulas, we know that $r+s=b$ and $r s=a-1$; thus, for $x=r$ and $y=s$, we have $f(f(r) f(s))+f(b)=f(a-1)=f(a)-1$.

From the equation, we can deduce $f(a)=f(b): f(f(r) f(s))=-1, f(f(r) f(s)+1)=0$, so $f(r) f(s)+1=1$, i.e., $f(r) f(s)=0$. We can assume that $s=1$, then $a=1 \cdot r+1$ and $b=r+1$, so $a=b$, which leads to a contradiction; the function is indeed injective.

Let $y=1-x$. Then $f(f(x) f(1-x))+f(1)=f(x(1-x))$,

$$
f(f(x) f(1-x))=f\left(x-x^{2}\right)
$$

By injectivity, $f(x) f(1-x)=x-x^{2}$.

Now let $y=-x$. Then $f(f(x) f(-x))+f(0)=f\left(-x^{2}\right)$,

$$
f(f(x) f(-x))-1=f\left(-x^{2}\right)
$$

$f(f(x) f(-x))=f\left(1-x^{2}\right)$.

By injectivity, $f(x) f(-x)=1-x^{2}$, so $f(x) f(1-x)-f(x) f(-x)=x-x^{2}-(1-x^{2})=x-1$. On the other hand,

$$
f(x) f(1-x)-f(x) f(-x)=f(x)(f(1-x)-f(-x))=f(x)
$$

Thus, $f(x)=x-1$ for all $x$. This is indeed a valid solution: $(x-1)(y-1)-1+x+y-1=xy-1$.

This was the solution when $f(0)=-1$, and its opposite, $f(x)=1-x$, is the solution when $f(0)=-1$.

Therefore, the functional equation has three solutions: $f(x)=0$, $f(x)=x-1$, and $f(x)=1-x$.