Let's determine all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that for any $x, y$, $f(x+y) + f(x) f(y) = x^{2} y^{2} + 2 x y$.
Solution. Let $x=y=0$:

\[
\begin{aligned}
f(0+0)+f(0) \cdot f(0) & =0^{2} \cdot 0^{2}+2 \cdot 0 \cdot 0 \\
f(0)^{2}+f(0) & =0 \\
f(0) \cdot(f(0)+1) & =0
\end{aligned}
\]

A product is zero if and only if one of its factors is zero, so $f(0)=0$ or $f(0)=-1$.

Let $y=0$ and $x$ be arbitrary:

\[
\begin{aligned}
f(x+0)+f(x) \cdot f(0) & =x^{2} \cdot 0^{2}+2 x \cdot 0, \\
f(x)+f(x) \cdot f(0)=f(x) \cdot(1+f(0)) & =0
\end{aligned}
\]

If $f(0)=0$, then the first factor of the product must be zero:

\[
f(x)=0
\]

However, this is not a solution for $x=y=1$:

\[
\begin{aligned}
f(2)+f(1)^{2} & =1^{2} \cdot 1^{2}+2 \cdot 1 \cdot 1 \\
0 & \neq 3
\end{aligned}
\]

Therefore, $f(0)=-1$.

Let $y=-x$:

\[
\begin{aligned}
f(x-x)+f(x) \cdot f(-x) & =x^{4}-2 x^{2} \\
f(0)+f(x) \cdot f(-x) & =x^{4}-2 x^{2} .
\end{aligned}
\]

Using the fact that $f(0)=-1$:

\[
\begin{aligned}
& f(x) \cdot f(-x)=x^{4}-2 x^{2}+1 \\
& f(x) \cdot f(-x)=\left(x^{2}-1\right)^{2}
\end{aligned}
\]

Let $x=1$:

\[
f(1) f(-1)=0 \text{. }
\]

A product is zero if and only if one of its factors is zero.

If $f(1)=0$, then let $y=1$ and $x$ be arbitrary:

\[
\begin{aligned}
f(x+1)+f(x) \cdot f(1) & =x^{2} \cdot 1^{2}+2 x \cdot 1 \\
f(x+1)+f(x) \cdot 0 & =x^{2}+2 x \\
f(x+1) & =x^{2}+2 x
\end{aligned}
\]

Thus, in this case, $f(x)=(x-1)^{2}+2(x-1)=x^{2}-1$.

If $f(-1)=0$, then let $y=-1$ and $x$ be arbitrary:

\[
\begin{aligned}
f(x-1)+f(x) \cdot f(-1) & =x^{2} \cdot(-1)^{2}+2 x \cdot(-1) \\
f(x-1)+f(x) \cdot 0 & =x^{2}-2 x \\
f(x-1) & =x^{2}-2 x
\end{aligned}
\]

Thus, in this case as well, $f(x)=(x+1)^{2}-2(x+1)=x^{2}-1$.

This solution is indeed correct:

\[
\begin{gathered}
f(x+y)+f(x) \cdot f(y)=(x+y)^{2}-1+\left(x^{2}-1\right)\left(y^{2}-1\right)= \\
=x^{2}+2 x y+y^{2}-1+x^{2} y^{2}-x^{2}-y^{2}+1=x^{2} y^{2}+2 x y
\end{gathered}
\]

Therefore, the only such function $f$ is $f(x)=x^{2}-1$.