The approximation function of $\frac{x}{y}$ Is there a approximation function of $$\frac{x}{y},$$ and the approximation function is in the form of $f(x) + f(y)$ or $f(x) - f(y)$. That's to say the approximation function can split $x$ and $y$.

Though the question is unspecific about what constitutes an "approximation", the answer appears to be "no".
As lulu notes in the comments, an approximation $\frac{x}{y} \approx f(x) + f(y)$ leads (for $x = y$) to
$$
1 = \frac{x}{x} \approx f(x) + f(x) = 2f(x)\quad\text{for all $x$.}
$$
Similarly, an approximation $\frac{x}{y} \approx f(x) - f(y)$ leads (for $x = y$) to
$$
1 = \frac{x}{x} \approx f(x) - f(x) = 0.
$$

From the other direction (i.e., starting with customary notions of approximation and seeing where they lead):
If $y_{0} \neq 0$, then for $|y - y_{0}| < |y_{0}|$ the geometric series gives the first-order approximation
\begin{align*}
  \frac{x}{y} &= \frac{x}{y_{0} + (y - y_{0})}
  = \frac{x}{y_{0}} \cdot \frac{1}{1 + (\frac{y - y_{0}}{y_{0}})} \\
  &= \frac{x}{y_{0}} \cdot \left[1 - \frac{y - y_{0}}{y_{0}} + \bigg(\frac{y - y_{0}}{y_{0}}\biggr)^{2} - \cdots\right] \\
  &\approx \frac{x}{y_{0}} - \frac{x(y - y_{0})}{y_{0}^{2}},
\end{align*}
which is not of the form you seek.
