Is there any functional equation $f(ab+cd)= f(a)+f(b)+f(c)+f(d)$?

I am looking for a real, continuous function that satisfies the functional equation
$$
f(ab+cd)= f(a)+f(b)+f(c)+f(d)
$$
where $a,b,c,d$ are real.

This is equivalent to a function satisfying these two functional equations:
$$
f(x+y)=f(x)+f(y)
$$
$$
f(xy)=f(x)+f(y)
$$
or Cauchy's functional equation and the logarithmic functional equation, respectively. I have the suspicion that the only the function $f(x) = 0$ satisfies this, but I am very confused as to how a solution may be found analytically. Thank you in advance.

EDIT: I saw in the comments that if f(x) is defined at zero the answer is trivial. If the domain were restricted to not include zero or even to not include all nonpositive numbers, would the answer change? If so, how? What if the function were discontinuous? Thank you again.

Answer

If $f$ has the positive numbers as its domain, it is zero regardless of continuity.

Note first that $f$ satisfies the relation $f(x+y)=f(xy)$ for all $x,y$ in its domain. Now, fix any two positive numbers $u$ and $v$. If $u^2> 4v$, then $u=x+y$ and $v=xy$ for $x,y=\frac{u \pm \sqrt{u^2-4v}}{2}$, and these $x,y$ are positive. So $f(u)=f(v)$.

Similarly, if $v^2 > 4u$, then $f(u)=f(v)$.

But if $u^2 \le 4v$ and $v^2\le4u$, then
$u^2\le4(2\sqrt{u})$, which implies that $u\le 4$. So $4^2 \ge 4u$, which means that $f(u)=f(4)$. Since $u$ and $v$ are symmetric, we also have $f(v)=f(4)$. So $f(u)=f(v)$.

Since $u$ and $v$ are arbitrary, $f$ is constant, which means that it is zero.