This would be a sort of complex-valued analogue of the Euclidean metric.
I would have thought that we could prove there isn't such an $f$ by using the fact that there isn't a continuous right inverse to $z\mapsto z^2$. But I can't find a proof or an example.
On
We have $f(x,1)^2=x^2+1$, and $f(\cdot,1)$ is continuous.
Note that since there is a holomorphic square root in a neighborhood of any complex number but zero, $f(\cdot,1)$ is continuous and holomorphic everywhere except at $i$ and $-i$. By Moreira, $f(\cdot,1)$ is holomorphic.
It vanishes at $x=i$, so $f(\cdot,1)^2=X^2+1$ should vanish with order $2$ at least at $i$. This is not the case, hence a contradiction and $f$ does not exist.
Mindlack's answer is good, but I found a nicer answer here in the meantime.
Suppose such an $f$ exists, and for $z\in\mathbb C$ let $x = \frac{z+1}{2}$ and $y=\frac{z-1}{2i}$. Then
$$f\left(\frac{z+1}{2},\frac{z-1}{2i}\right)^2 = \left(\frac{z+1}{2}\right)^2 + \left(\frac{z-1}{2i}\right)^2=\frac{z^2+2z+1}4-\frac{z^2-2z+1}4=z$$
so $z\mapsto f\left(\frac{z+1}{2},\frac{z-1}{2i}\right)$ is a continuous right inverse to $z\mapsto z^2$, contradiction.