4: Let $F(x, y)=(x-y)^{2}+\left(\frac{x}{3}+\frac{3}{y}\right)(y \neq 0)$. Then the minimum value of $F(x, y)$ is $\qquad$
4. $\frac{18}{5}$.

Let $P\left(x, \frac{x}{3}\right), Q\left(y,-\frac{3}{y}\right)$.
Then $F(x, y)=|P Q|^{2}$, where $P$ is a moving point on the line $l_{1}: y=\frac{1}{3} x$, and $Q$ is a moving point on the curve $y=-\frac{3}{x}$.
Suppose the line $l_{2}: y=\frac{1}{3} x+m$ is tangent to the curve $y=-\frac{3}{x}$.
Then $x^{2}+3 m x+9=0$.
Let $\Delta=9 m^{2}-36=0 \Rightarrow m= \pm 2$.
By symmetry, we can take $l_{2}: y=\frac{1}{3} x+2$.
Since the distance between the lines $l_{1}$ and $l_{2}$ is
$$
d=\frac{|0-2|}{\sqrt{1+\frac{1}{9}}}=\frac{6}{\sqrt{10}},
$$

Therefore, the minimum value of $|P Q|$ is $d=\frac{6}{\sqrt{10}}$.
Hence, the minimum value of $F(x, y)$ is $\frac{36}{10}=\frac{18}{5}$.